CBSE Class 11 Maths Revision Notes Chapter 4

Q.1 Prove by the method of induction that every even power of every odd integer greater than 1 when divided by 8 leaves the remainder 1.

Ans

$$\begin{gathered} \mathrm{First}\hspace{0.17em}\mathrm{odd}\mathrm{integer}>1\mathrm{is}3, \\ \text{general Odd\hspace{0.17em}integer is\hspace{0.17em}}\left(2\mathrm{r}+1\right) \\ \text{To show} \\ {\left(2\mathrm{r}+1\right)}^{2\mathrm{n}}=8\mathrm{m}+1,\hspace{0.17em}\mathrm{m}\in \mathrm{N} \\ \text{i.e.\hspace{0.17em}}{\left(2\mathrm{r}+1\right)}^{2\mathrm{n}}–1\text{\hspace{0.17em}is divisible by\hspace{0.17em}}8 \\ \text{Let,\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}:\hspace{0.17em}{\left(2\mathrm{x}+1\right)}^{2\mathrm{n}}–1=4{\mathrm{r}}^2+4\mathrm{r} \\ =4\mathrm{r}\left(\mathrm{r}+1\right) \\ \text{As\hspace{0.17em}}\mathrm{r}\left(\mathrm{r}+1\right)\text{is always even} \\ \therefore \hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(1\right)\text{\hspace{0.17em}is true} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is true} \\ \mathrm{P}\left(\mathrm{k}\right)=2\mathrm{r}+1^{2\mathrm{k}}–1\text{\hspace{0.17em}is divisible by\hspace{0.17em}}8 \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true} \\ \text{i.e\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=2\mathrm{r}+{12}^{(\mathrm{k}+1)}–1\text{\hspace{0.17em}is divisible by\hspace{0.17em}}8 \\ \mathrm{P}\left(\mathrm{k}+1\right)={\left(2\mathrm{r}+1\right)}^{2\mathrm{k}+2}–1 \\ ={\left(2\mathrm{r}+1\right)}^{2\mathrm{k}}.{\left(2\mathrm{r}+1\right)}^2–1 \\ =\left(8\mathrm{m}+1\right)\left(8\mathrm{p}+1\right)–1 \\ =64\mathrm{mp}+8\mathrm{m}+8\mathrm{p}+1–1 \\ =8\left[8\mathrm{mp}+\mathrm{m}+\mathrm{p}\right]\text{\hspace{0.17em}Which is divisible by\hspace{0.17em}}8 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is true} \\ \text{Hence by principle of mathematical induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true . for all n\hspace{0.17em}∈\hspace{0.17em}N.} \end{gathered}$$

Q.2

$\mathbf{\text{Prove that}}{(1+X)}^{\mathbf{n}}\mathbf{\ge }(1+\mathrm{nx})\mathbf{\text{for all natural number}}\mathbf{n}\mathbf{,}\mathbf{\text{where}}\mathbf{x}\mathbf{>}\mathbf{–}\mathbf{1}$

Ans

$$\begin{gathered} Let\hspace{0.17em}\mathrm{P}\left(\mathrm{n}\right)=(1+\mathrm{x}{)}^{\mathrm{n}}\ge (1+\mathrm{nx}) \\ \Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)\hspace{0.17em}:\hspace{0.17em}(1+\mathrm{x}{)}^{\mathrm{k}}\ge (1+\mathrm{kx}) \\ \text{We Shall Prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True for}\mathrm{x}>–1 \\ \text{Now}\mathrm{P}\left(\mathrm{k}+1\right)\hspace{0.17em}:\hspace{0.17em}(1+\mathrm{x}{)}^{\mathrm{k}+1} \\ (1+\mathrm{x}{)}^{\mathrm{k}+1}\ge \left(1+\mathrm{x}+\mathrm{kx}+{\mathrm{kx}}^2\right) \\ \text{Here}\mathrm{K}\text{is a natural number and}{\mathrm{x}}^2\ge 0\text{so that}{\mathrm{kx}}^2\ge 0 \\ \therefore \left(1+\mathrm{x}+\mathrm{kx}+{\mathrm{kx}}^2\right)\ge \left(1+\mathrm{x}+\mathrm{kx}\right) \\ \text{And so we obtain} \\ (1+\mathrm{x}{)}^{\mathrm{k}+1}\ge (1+\mathrm{x}+\mathrm{kx}\left) \\ \right(1+\mathrm{x}{)}^{\mathrm{k}+1}\ge [1+(1+\mathrm{k}\left)\mathrm{x}\right] \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right) \\ \text{Hence by principle of mathematical indention}\mathrm{P}\left(\mathrm{n}\right)\text{is true for where\hspace{0.17em}n\hspace{0.17em}∈\hspace{0.17em}N.} \end{gathered}$$

Q.3

$\mathbf{\text{Prove that}}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{>}\frac{{\mathbf{n}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{,}\mathbf{\hspace{0.17em}}\mathbf{n}\mathbf{\in }\mathbf{N}$

Ans

$$\begin{gathered} \text{Let}\mathrm{P}\left(\mathrm{n}\right)=1^2+2^2+\dots +{\mathrm{n}}^2>\frac{{\mathrm{n}}^3}{3} \\ \mathrm{P}\left(1\right)=1>\frac{1}{3} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=1^2+2^2+\dots .+{\mathrm{k}}^2>\frac{{\mathrm{k}}^3}{3} \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{i.e}\mathrm{P}\left(\mathrm{k}+1\right)=1^2+2^2+\dots ..+{\mathrm{k}}^2+{\left(\mathrm{k}+1\right)}^2>\frac{{\left(\mathrm{k}+1\right)}^3}{3} \\ \text{Now}\mathrm{P}\left(\mathrm{k}+1\right)=\left(1+2^2+\dots .+{\mathrm{k}}^2\right)+{\left(\mathrm{k}+1\right)}^2>\frac{{\mathrm{k}}^3}{3}+{\left(\mathrm{k}+1\right)}^2 \\ =\frac{1}{3}\left[{\mathrm{k}}^3+3{\mathrm{k}}^2+6\mathrm{k}+3\right] \\ =\frac{1}{3}\left[(\mathrm{k}+1{)}^3+3\mathrm{k}+2\right]>\frac{1}{3}(\mathrm{k}+1{)}^3 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle of mathematical induction is true. For all\hspace{0.17em}n\hspace{0.17em}}\in \mathrm{N}\text{.} \end{gathered}$$

Q.4 Prove that: 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24.

Ans

$$\begin{gathered} \text{Let\hspace{0.17em}P\hspace{0.17em}}\left(\mathrm{n}\right)=2.7^{\mathrm{n}}+3.5^{\mathrm{n}}–5 \\ \mathrm{P}\left(1\right)=2.7^1+3.5^1–5=14+15–5=24 \\ \Rightarrow \mathrm{P}\left(1\right)\text{\hspace{0.17em}is true} \\ \text{Assuming\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right)\text{\hspace{0.17em}is true} \\ \mathrm{P}\left(\mathrm{k}\right)=2.7^{\mathrm{k}}+3.5^{\mathrm{k}}–5\text{is divisible by}24 \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true} \\ \text{i.e.\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=2.7^{\mathrm{k}+1}+3.5^{\mathrm{k}+1}–5\text{\hspace{0.17em}is divisible by\hspace{0.17em}}24 \\ 2.7^{\mathrm{k}+1}+3.5^{\mathrm{k}+1}–5=24\mathrm{g}\left(\mathrm{let}\right)...\left(1\right) \\ \mathrm{P}\left(\mathrm{k}+1\right)=2.7^{\mathrm{k}+1}+3.5^{\mathrm{k}+1}–5 \\ =2.7^{\mathrm{k}}.7+3.5^{\mathrm{k}}.5–5 \\ =7\left[2.7^{\mathrm{k}}+3.5^{\mathrm{k}}–5–3.5^{\mathrm{k}}+5\right]+3.5^{\mathrm{k}}.5–5 \\ =7\left[24\mathrm{g}–3.5^{\mathrm{k}}–5–3.5^{\mathrm{k}}–5\right]+15.5^{\mathrm{k}}–5\hspace{0.17em}\left[\text{From equation\hspace{0.17em}}\left(1\right)\right] \\ =7\times 24\mathrm{g}–21.5^{\mathrm{k}}+35+15.5^{\mathrm{k}}–5 \\ =7\times 24\mathrm{g}–6.5^{\mathrm{k}}+30 \\ =7\times 24\mathrm{g}–6\left(5^{\mathrm{K}}–5\right) \\ =7\times 24\mathrm{g}–6\left(4\mathrm{P}\right)\left[âˆ\mu \hspace{0.17em}{5}^{\mathrm{k}}–5\text{\hspace{0.17em}is a multiple of\hspace{0.17em}}{45}^{\mathrm{k}}–5=4\mathrm{P}\right] \\ =7\times 24\mathrm{g}–24\mathrm{P} \\ \text{Which is divisible by\hspace{0.17em}}24 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is true} \\ \text{Hence by principle of mathematical induction is true. For all\hspace{0.17em}n ∈ N} \end{gathered}$$

Q.5

$$\begin{gathered} \mathbf{\text{Prove that:\hspace{0.17em}}}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{)}\mathbf{\text{\hspace{0.17em}…..\hspace{0.17em}}}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{\text{by using principle of mathematical induction.}} \end{gathered}$$

Ans

$$\begin{gathered} \text{Let}\mathrm{P}\left(\mathrm{n}\right)=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)⋯·\left(1+\frac{1}{\mathrm{n}}\right)=\left(\mathrm{n}+1\right) \\ \mathrm{P}\left(1\right)=\left(1+\frac{1}{1}\right)=\left(1+1\right) \\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2=2 \\ \Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots \left(1+\frac{1}{\mathrm{k}}\right)=\left(\mathrm{k}+1\right) \\ \text{We shall prove\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \\ \text{i.e.,\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=\left(\mathrm{k}+2\right) \\ \text{Now\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)⋯\left(1+\frac{1}{\mathrm{k}}\right)\left(1+\frac{1}{\mathrm{k}+1}\right) \\ =\left(\mathrm{k}+1\right)\left(1+\frac{1}{\mathrm{k}+1}\right)=\left(\cancel{\mathrm{k}+1}\right)\frac{(\mathrm{k}+1+1)}{\left(\cancel{\mathrm{k}+1}\right)}=\left(\mathrm{k}+2\right) \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by mathematical induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true for all\hspace{0.17em}}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$

Q.6 Prove that 7ⁿ – 3ⁿ is divisible by 4 for all n ∈ N.

Ans

$$\begin{gathered} Let\mathrm{P}\left(\mathrm{n}\right)=7^{\mathrm{n}}{3}^{\mathrm{n}} \\ \mathrm{P}\left(1\right)=7–3=4\text{which is divisible by}4 \\ \Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=7^{\mathrm{k}}–3^{\mathrm{k}}\text{is divisible by}4 \\ \text{Let}{7}^{\mathrm{k}}–3^{\mathrm{k}}=4\mathrm{d}...\left(1\right) \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{i.e.}\mathrm{P}\left(\mathrm{k}+1\right)=7^{\mathrm{k}+1}–3^{\mathrm{k}+1}\text{is divisible by}4 \\ \text{Now\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=7^{\mathrm{k}+1}–3^{\mathrm{k}+1}=7^{\mathrm{k}}.7–3^{\mathrm{k}}.3 \\ =7^{\mathrm{k}}.7–7.3^{\mathrm{k}}+7.3^{\mathrm{k}}–3.3^{\mathrm{k}} \\ =7\left(7^{\mathrm{k}}–3^{\mathrm{k}}\right)+\left(7–3\right).3^{\mathrm{k}} \\ =7\left(4\mathrm{d}\right)+4.3^{\mathrm{k}} \\ \text{Hence}\mathrm{P}\left(\mathrm{k}+1\right)\text{is divisible by}4 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle of mathematical induction is true. For all\hspace{0.17em}n\hspace{0.17em}∈\hspace{0.17em}N} \end{gathered}$$

Q.7

$\mathbf{\text{Prove that\hspace{0.17em}}}\frac{\mathbf{1}}{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{4}}\mathbf{+}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{)}}{\mathbf{4}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}\mathbf{\text{\hspace{0.17em}for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}$

Ans

$$\begin{gathered} Let\hspace{0.17em}\mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}:\frac{1}{1.2·3}+\frac{1}{2.3.4}+\dots \dots +\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\frac{\mathrm{n}(\mathrm{n}+3)}{4(\mathrm{n}+1)(\mathrm{n}+2)} \\ \mathrm{P}\left(1\right)=\frac{1}{1.2.3}=\frac{1(1+3)}{4(1+1)(1+2)}\Rightarrow \frac{1}{6}=\frac{\cancel{4}}{\cancel{4}.2.3} \\ \text{⇒\hspace{0.17em}}\frac{1}{6}=\frac{1}{6}\Rightarrow \mathrm{P}\left(1\right)\text{\hspace{0.17em}is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\dots ..+\frac{1}{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)} \\ =\frac{\mathrm{k}(\mathrm{k}+3)}{4(\mathrm{k}+1)(\mathrm{k}+2)} \\ \text{We shall prove\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \\ \text{i.e}\mathrm{P}\left(\mathrm{k}+1\right)=\frac{(\mathrm{k}+1)(\mathrm{k}+4)}{4(\mathrm{k}+2)(\mathrm{k}+3)} \\ \text{Now\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=\frac{\mathrm{k}(\mathrm{k}+3)}{4(\mathrm{k}+1)(\mathrm{k}+2)}+\frac{1}{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)} \\ =\frac{\mathrm{k}(\mathrm{k}+3{)}^2+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)} \\ =\frac{\mathrm{k}\left({\mathrm{k}}^2+6\mathrm{k}+9\right)+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)} \\ =\frac{{\mathrm{k}}^3+6{\mathrm{k}}^2+9\mathrm{k}+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)} \\ =\frac{(\mathrm{k}+1{)}^2(\mathrm{k}+4)}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+4)}=\frac{(\mathrm{k}+1)(\mathrm{k}+4)}{4(\mathrm{k}+2)(\mathrm{k}+3)} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \\ \text{Hence by principle of mathematical induction is true. For all\hspace{0.17em}}\mathrm{n}\in \mathrm{N} \end{gathered}$$

Q.8 Prove that 1.2 + 2.3 + 3.4 + … + n(n +1) = n( n +1)(n + 2)/3.

Ans

$$\begin{gathered} \text{Let\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}\hspace{0.17em}:1.2+2.3+3.4+\dots +\mathrm{n}.\left(\mathrm{n}+1\right)=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3} \\ \mathrm{P}\left(1\right)=1.2=\frac{1(1+1)(1+2)}{3}\Rightarrow 2=\frac{2\times 3}{3}\Rightarrow 2=2 \\ \Rightarrow \mathrm{P}\left(1\right)\text{\hspace{0.17em}is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=1.2+2.3+3.4+\dots \dots +\mathrm{k}=\frac{\mathrm{k}(\mathrm{k}+1)\left(\mathrm{k}+2\right)}{3} \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is\hspace{0.17em}True} \\ \text{i.e\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=1.2+2.3+3.4+\dots \dots +\mathrm{k}.\left(\mathrm{k}+1\right)+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right) \\ =\frac{(\mathrm{k}+1)\mathrm{k}+2\left)\right(\mathrm{k}+2)}{3} \\ \mathrm{P}\left(\mathrm{k}+1\right)=\frac{\mathrm{k}(\mathrm{k}+1)\mathrm{k}+2)}{3}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right) \\ =\frac{\mathrm{k}(\mathrm{k}+1)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{3} \\ =\frac{(\mathrm{k}+1)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{3} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \end{gathered}$$

Q.9 Prove that : 2ⁿ > n for all positive integers n.

Ans

$$\begin{gathered} \text{When}\mathrm{n}=1, \\ {2}^1>1\text{. Hence}\mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right):2^{\mathrm{k}}>\mathrm{k} \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{i.e.,}\mathrm{P}\left(\mathrm{k}+1\right)=2^{\mathrm{k}+1}>\mathrm{k}+1 \\ \mathrm{P}\left(\mathrm{k}+1\right)=2^{\mathrm{k}+1}=2^{\mathrm{k}}·2^1=2^{\mathrm{k}}·2 \\ \Rightarrow 2^{\mathrm{k}}.2>2\mathrm{k}=\mathrm{k}+\mathrm{k}>\mathrm{k}+1 \\ {2}^{\mathrm{k}+1}>2\mathrm{k}>\mathrm{k}+1 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true, for all}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$

Q.10

$\mathbf{Prove}\mathbf{}\mathbf{that}\mathbf{}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{\text{\hspace{0.17em}for all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}$

Ans

$$\begin{gathered} \text{Let,\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)=1^2+2^2+3^3+\dots +{\mathrm{n}}^2 \\ \mathrm{P}\left(1\right)=1^2=1^2\Rightarrow 1=1\Rightarrow \mathrm{P}\left(1\right)\text{is true} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=1^2+2^2+3^3+\dots \dots ..+{\mathrm{k}}^2=\frac{\mathrm{k}(\mathrm{k}+1)(2\mathrm{k}+1)}{6} \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \\ \mathrm{P}\left(\mathrm{k}+1\right)=1^2+2^2+3^3+\dots \dots +{\mathrm{k}}^2+{\left(\mathrm{k}+1\right)}^2 \\ =\frac{(\mathrm{k}+1)(\mathrm{k}+2)(2\mathrm{k}+3)}{6} \\ \mathrm{P}\left(\mathrm{k}+1\right)=\frac{\mathrm{k}(\mathrm{k}+1)(2\mathrm{k}+1)}{6}+{\left(\mathrm{k}+1\right)}^2 \\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(2\mathrm{k}+1\right)+6{\left(\mathrm{k}+1\right)}^2}{6} \\ =\frac{(\mathrm{k}+1)\left(2{\mathrm{k}}^2+\mathrm{k}+6\mathrm{k}+6\right)}{6} \\ =\frac{(\mathrm{k}+1)\left(2{\mathrm{k}}^2+7\mathrm{k}+6\right)}{6} \\ =\frac{(\mathrm{k}+1)\left(2{\mathrm{k}}^2+4\mathrm{k}+3\mathrm{k}+6\right)}{6} \\ =\frac{(\mathrm{k}+1)\left[2\mathrm{k}\right(\mathrm{k}+2)+3(\mathrm{k}+2\left)\right]}{6} \\ =\frac{(\mathrm{k}+1)(2\mathrm{k}+3)(\mathrm{k}+2)}{6}=\frac{(\mathrm{k}+1)(\mathrm{k}+2)(2\mathrm{k}+3)}{6} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is\hspace{0.17em}True} \\ \text{Hence, by principle of mathematical induction\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is\hspace{0.17em}true.\hspace{0.17em}For\hspace{0.17em}all\hspace{0.17em}}\mathrm{n}\in \mathrm{N} \end{gathered}$$

Q.11 Prove that: 1 + 3 + 5 +…+ (2n – 1) = n^2.

Ans

$$\begin{gathered} \text{Let,}\mathrm{P}\left(\mathrm{n}\right)=1+3+5+\dots ..+\left(2\mathrm{n}–1\right)={\mathrm{n}}^2 \\ \mathrm{P}\left(1\right)=1=1^2\Rightarrow 1=1\Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=1+3+5+\dots \dots .+\left(2\mathrm{k}–1\right)={\mathrm{k}}^2 \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{i.e.,}\mathrm{P}\left(\mathrm{k}+1\right)=1+3+5\dots \dots +\left(2\mathrm{k}–1\right)+\left(2\mathrm{k}+1\right)={\left(\mathrm{k}+1\right)}^2 \\ \text{Now}\mathrm{P}\left(\mathrm{k}+1\right)={\mathrm{k}}^2+2\mathrm{k}+1 \\ ={\left(\mathrm{k}+1\right)}^2 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle of mathematical induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true. For all}\mathrm{n}\in \mathrm{N} \end{gathered}$$

Q.12 Prove that: 1 + 2 + 3 +…+ n = n(n + 1)/2 for all n ∈ N.

Ans

$$\begin{gathered} \text{Let}\mathrm{P}\left(\mathrm{n}\right)=1+2+3+\dots \dots +\mathrm{n}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2} \\ \mathrm{P}\left(1\right)=1=\frac{1\left(1+1\right)}{2}\Rightarrow 1=1\Rightarrow \mathrm{P}\left(1\right)\text{is True.} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True.} \\ \mathrm{P}\left(\mathrm{k}\right)=1+2+3+\dots \dots +\mathrm{k}=\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2} \\ \text{Now}\mathrm{P}\left(\mathrm{k}+1\right)=1+2+3+\dots \dots +\mathrm{k}+\left(\mathrm{k}+1\right)=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{2} \\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}+\left(\mathrm{k}+1\right) \\ =\frac{{\mathrm{k}}^2+\mathrm{k}+2\mathrm{k}+2}{2} \\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)+2\left(\mathrm{k}+1\right)}{2} \\ =\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{2} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle of mathematical induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true for all}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$

Q.13 Suppose P(n): n(n+1)(n+2) is divisible by 6. Show that P(1), P(2) and P(3) are true.

Ans

Here
P(n) = n(n + 1)(n + 2)
P(1) = 1(1 + 1)(1 + 2) = 6 divisible by 6
P(2) = 2(2 + 1)(2 + 2) = 24 divisible by 6
P(1) = 3(3 + 1)(3 + 2) = 60 divisible by 6
Hence P(1), P(2) and P(3) are true.

Q.14 Let P(n)be statement 3n ≥ n!. where n is a natural number, then show that P(n) is true for n =1 and 2 .

Ans

P(1): 3 > 1! = 3 > 1 , which is true.
P(2): 6 > 2! = 6 > 2, which is true.

Q.15 If P(n) be the statement aⁿ + abⁿ is divisible by a, show that P(1) and P(2) are true.

Ans

P(n) : aⁿ + abⁿ is divisible by a
P(1) : a + b which is is divisible by a.
P(2): a² + ab² = a(a + b²), which is is divisible by a.
Hence P(1) and P(2) are true.

Q.16

$\mathbf{\text{If\hspace{0.17em}}}{\mathbf{A}}^{\mathbf{n}}\mathbf{=}\left[\begin{array}{cc}1& \text{na}\\ 0& 1\end{array}\right]\mathbf{,}\mathbf{\text{\hspace{0.17em}find\hspace{0.17em}}}\mathbf{A}\mathbf{\text{\hspace{0.17em}and\hspace{0.17em}}}{\mathbf{A}}^{\mathbf{2}}$

Ans

$$\begin{gathered} {\mathrm{A}}^{\mathrm{n}}=\left[\begin{array}{cc}1& \text{na}\\ 0& 1\end{array}\right], \\ \text{Substitute}\mathrm{n}=1,\text{we get} \\ \mathrm{A}=\left[\begin{array}{ll}1& \mathrm{a}\\ 0& 1\end{array}\right]\text{and} \\ {\mathrm{A}}^2=\left[\begin{array}{ll}1& \mathrm{a}\\ 0& 1\end{array}\right]\times \left[\begin{array}{ll}1& \mathrm{a}\\ 0& 1\end{array}\right]=\left[\begin{array}{ll}1& 2\mathrm{a}\\ 0& 1\end{array}\right]. \end{gathered}$$

Q.17 If P(n) : 7²ⁿ + 2^{3n – 3} 3^{n – 1} is divisible by 25, show that P(1) and P(2) are true.

Ans

P(n) : 7²ⁿ + 2^{3n – 3} 3^{n – 1}
P(1) : 7² + 2^{3 – 3} 3 ^{1 – 1} = 49 + 2⁰ 3 ⁰ = 49 + 1 = 50
P(2) : 7^2.2 + 2^{3.2 – 3} 3 ^{2 – 1} = 2401 + 24 = 2425
Which is divisible by 25 only.

Q.18 1³ + 2³ + 3³ + … + 1000³ = K² . Find the value of K.

Ans

We have,
1³ + 2³ + 3³ + … + 1000³ = K² , Σn³ = K² from n = 1 to 1000.
{1000(1000 + 1)/2}² = K²
Therefore, K = 250250.

Q.19 Show that a²ⁿ – b²ⁿ (a, b are distinct rational numbers) is divisible by a – b for n = 1, 2 and 3.

Ans

For n = 1,
a²ⁿ – b²ⁿ = a – b which is divisible by a – b
For n = 2,
a²ⁿ – b²ⁿ = a⁴ – b⁴ = (a² – b²)(a² + b²) = (a – b)(a + b)(a² + b²) Which is divisible by a – b
For n = 3
a²ⁿ – b²ⁿ = a⁶ – b⁶ = (a² – b²)(a² + b² + ab) = (a – b)(a + b)(a² + b² + ab) Which is divisible by a – b.

Q.20 What do you mean by Principle of mathematical induction?

Ans

To prove certain results or statements that are formulated in the terms of n with a well-suited principle that is user-based on the specific technique, where n is a positive integer, is known as the principle of mathematical induction.

Q.21 What is the main application of Principle of Mathematical Induction?

Ans

The principal of mathematical induction is used to prove the mathematical truth in the form of well defined statement or formula.

Q.22 What is the main drawback of Principle of Mathematical Induction?

Ans

The principal of mathematical Induction cannot be used to –

(i) establish a formula.
(ii) solve a equation.
(iii) calculate or operate any mathematical operation.
(iv) any operation other than proof.

Q.23 What do you mean by Principle of Mathematical Induction method?

Ans

In Principal of Mathematical Induction method, we proceed from particular cases to general cases.

Q.24 What do you mean by Principle of Mathematical Deduction method?

Ans

In Principle of Mathematical Deduction method, we proceed from general cases to particular cases.

Q.25 What is the basic assumption in Principle of Mathematical Induction?

Ans

First, we test P(n) for n = 1. If it is true, then we let n = k and on the basis of this, we test for n = k + 1.

Q.26 Write the algorithm for Principle of Mathematical Induction.

Ans

Step-I Obtain the statement P(n).
Step-II Test P(n) is true for n=1.
Step-III Let P(n) is true for n=k.
Step-IV Test P(n) is true for n=k+1 by using step III.

Q.27

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{\text{\hspace{0.17em}for all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N} \\ \mathbf{\text{then\hspace{0.17em}prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):1^2+2^2+3^2+\dots \dots ..+{\mathrm{n}}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6} \\ \mathrm{LHS}=1^2=1\left[\text{take first term, for}\mathrm{n}=1\right] \\ \text{Put}\mathrm{n}=1\text{in RHS} \\ \mathrm{RHS}=\frac{1\left(1+1\right)\left(2\times 1+1\right)}{6}=\frac{2\times 3}{6}=1 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is true} \end{gathered}$$

Q.28

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N} \\ \mathbf{\text{then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):1^2+2^2+3^2+\dots \dots +{\mathrm{n}}^2=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6} \\ \mathrm{LHS}=1^2+2^2=1+4=5\left[\text{take first two terms, for}\mathrm{n}=2\right] \\ \text{Put}\mathrm{n}=2\text{in RHS} \\ \mathrm{RHS}=\frac{2\left(2+1\right)\left(2\times 2+1\right)}{6}=\frac{2\times 3\times 5}{6}=5 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(2\right)\text{is true.} \end{gathered}$$

Q.29

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{n}\mathbf{·}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\left[\frac{n(n+1)(n+2)}{3}\right] \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{\text{\hspace{0.17em}then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):1.2+2.3+3.4+\dots ..+\mathrm{n}·\left(\mathrm{n}+1\right)=\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{3}\right] \\ \mathrm{LHS}=1.2=2\left[\text{take first term, for}\mathrm{n}=1\right] \\ \text{Put}\mathrm{n}=1\text{in}\mathrm{RHS} \\ \mathrm{RHS}=\frac{1\left(1+1\right)\left(1+2\right)}{3}=\frac{1\times 2\times 3}{3}=2 \\ \Rightarrow \text{LHS}=\text{RHS} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is true.} \end{gathered}$$

Q.30

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{n}\mathbf{·}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{\left[}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}{\mathbf{3}}\mathbf{\right]} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{\text{\hspace{0.17em}then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\text{is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):1.2+2.3+3.4+\dots ..+\mathrm{n}·\left(\mathrm{n}+1\right)=\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{3}\right] \\ \text{LHS}=1.2+2.3=2+6=8\left[\mathrm{take}\mathrm{first}\mathrm{two}\mathrm{term},\mathrm{for}\mathrm{n}=2\right] \\ \text{Put}\mathrm{n}=2\text{in RHS} \\ \mathrm{RHS}=\frac{2\left(2+1\right)\left(2+2\right)}{3}=\frac{2\times 3\times 4}{3}=8 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(2\right)\text{is true.} \end{gathered}$$

Q.31

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{)}\mathbf{\dots }\mathbf{·}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{\text{\hspace{0.17em}then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)⋯·\left(1+\frac{1}{\mathrm{n}}\right)=\left(\mathrm{n}+1\right) \\ \mathrm{LHS}=\left(1+\frac{1}{1}\right)=2\left[\text{take first term, for}\mathrm{n}=1\right] \\ \text{Put}\mathrm{n}=1\text{in RHS.} \\ \text{RHS}=\left(1+1\right)=2 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is true.} \end{gathered}$$

Q.32

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{)}\mathbf{}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{\text{\hspace{0.17em}then, prove that}}\mathbf{P}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\text{is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots ·\left(1+\frac{1}{\mathrm{n}}\right)=\left(\mathrm{n}+1\right) \\ LHS=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)=2\times \frac{3}{2}=3\left[\mathrm{take}\mathrm{first}\mathrm{two}\mathrm{terms},\mathrm{for}\mathrm{n}=2\right] \\ \text{Put}\mathrm{n}=2\text{in RHS} \\ \text{RHS}=(2+1)=3 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(2\right)\text{is true.} \end{gathered}$$