Exercise 4.4 - Chapter 4 - Combinatorics and Mathematical Induction - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 4.4
Question 1.
By the principle of mathematical induction, prove that, for $n \geq 1$
Thus $\mathrm{P}(\mathrm{K})$ is true $\Rightarrow(k+1)$ is true.
Hence by principle of mathematical induction, $P(n)$ is true for all $n \in N$.
$
1^3+2^3+3^3+\ldots+n^3=\left(\frac{n(n+1)}{2}\right)^2
$
Solution:
$
\mathrm{P}(n)=1^3+2^3+3^3+\ldots+n^3=\left(\frac{n(n+1)}{2}\right)^2
$
For $n=1$
$
P(1)=1=\left[\frac{1(1+1)}{2}\right]^2 \Rightarrow 1=1
$
$\therefore \mathrm{P}(1)$ is true
Let $\mathrm{P}(n)$ be true for $n=k$
$
\begin{aligned}
\therefore \mathrm{P}(k) & =1^3+2^3+3^3+\ldots+k^3 \\
& =\left[\frac{k(k+1)}{2}\right]^2
\end{aligned}
$
For $n=k+1$
$
\begin{aligned}
\mathrm{P}(k+1) & =1^3+2^3+3^3+\ldots+k^3+(k+1)^3 \\
& =\left[\frac{k(k+1)}{2}\right]^2+(k+1)^3 \\
& =(k+1)^2\left[\frac{k^2}{4}+k+1\right] \\
& =(k+1)^2\left[\frac{k^2+4 k+4}{4}\right] \\
& =\frac{(k+1)^2(k+2)^2}{4}=\left[\frac{(k+1)(k+2)}{2}\right]^2
\end{aligned}
$
[Using (i)]
$\therefore \mathrm{P}(\mathrm{k}+1)$ is true.

Thus $\mathrm{P}(\mathrm{K})$ is true $\Rightarrow(k+1)$ is true.
Hence by principle of mathematical induction, $P(n)$ is true for all $n \in N$.
Question 2.
By the principle of mathematical induction, prove that, for $n>1$
$
1^2+3^2+5^2+\ldots+(2 n-1)^2=\frac{n(2 n-1)(2 n+1)}{3}
$
Solution:
Let $\mathrm{P}(n)=1^2+3^2+5^2+\ldots+(2 n-1)^2=\frac{n(2 n-1)(2 n+1)}{3}$
For $n=1$
$
\begin{array}{rlrl}
\text { For } n=1 & \mathrm{P}(1) & =(2 \times 1-1)^2=\frac{1(2 \times 1-1)(2 \times 1+1)}{3} \\
\Rightarrow \quad 1 & =\frac{1 \times 1 \times 3}{3}
\end{array}
$
$\therefore \mathrm{P}(1)$ is true
Let $P(n)$ be true for $n=k$
$
\therefore \mathrm{P}(k)=1^2+3^2+5^2+\ldots+(2 k-1)^2=\frac{k(2 k-1)(2 k+1)}{3}
$
For $n=k+1$
$
\begin{aligned}
\mathrm{RHS} & =\frac{(k+1)(2 k+1)(2 k+3)}{3} \\
\mathrm{LHS} & =\frac{k(2 k-1)(2 k+1)}{3}+(2 k+1)^2 \text { (Using (i)] }
\end{aligned}
$

$
\begin{aligned}
& =(2 k+1)\left[\frac{k(2 k-1)}{3}+(2 k+1)\right] \\
& =(2 k+1)\left[\frac{2 k^2-k+6 k+3}{3}\right] \\
& =\frac{(2 k-1)\left(2 k^2+5 k+3\right)}{3}=\frac{(k+1)(k+1)(2 k+3)}{3} \\
& =\frac{(k+1)(2 k+1)(2 k+3)}{3}
\end{aligned}
$
$\therefore \mathrm{P}(\mathrm{k}+1)$ is true
Thus $\mathrm{P}(\mathrm{k})$ is true $\Rightarrow \mathrm{P}(\mathrm{k}+1)$ is true. Hence by principle of mathematical induction, $\mathrm{P}(\mathrm{k})$ is true for all $\mathrm{n} \in$ N.
Question 3.
Prove that the sum of the first $n$ non-zero even numbers is $n^2+n$.
Solution:
Let $\mathrm{P}(\mathrm{n}): 2+4+6+\ldots+2 \mathrm{n}=\mathrm{n}^2+\mathrm{n}, \forall \mathrm{n} \in N$

Step 1:
$P(1): 2=1^2+1=2$ which is true for $P(1)$
Step 2:
$P(k): 2+4+6+\ldots+2 \mathrm{k}=\mathrm{k}^2+\mathrm{k}$. Let it be true.
Step 3:
$
\begin{aligned}
& \mathrm{P}(\mathrm{k}+1): 2+4+6+\ldots+2 \mathrm{k}+(2 \mathrm{k}+2) \\
& =\mathrm{k}^2+\mathrm{k}+(2 \mathrm{k}+2)=\mathrm{k}^2+3 \mathrm{k}+2 \\
& =\mathrm{k}^2+2 \mathrm{k}+\mathrm{k}+1+1 \\
& =(\mathrm{k}+1)^2+(\mathrm{k}+1)
\end{aligned}
$
Which is true for $P(k+1)$
So, $P(k+1)$ is true whenever $P(k)$ is true.
Question 4.
By the principle of Mathematical induction, prove that, for $n \geq 1$.
$
1.2+2.3+3.4+\ldots+n .(n+1)=\frac{n(n+1)(n+2)}{3}
$
Solution:
Let $\mathrm{P}(n)=1.2+2.3+3.4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$
For $n=1$
$
\begin{aligned}
& \qquad \mathrm{P}(1)=1(1+1)=\frac{1(1+1)(1+2)}{3} \\
& \Rightarrow 2=2 \\
& \therefore \mathrm{P}(1) \text { is true } \\
& \text { Let } \mathrm{P}(n) \text { be true for } n=k \\
& \therefore \quad \mathrm{P}(k)=1.2+2.3+3.4+\ldots+k(k+1) \\
& =\left[\frac{k(k+1)(k+2)}{3}\right]
\end{aligned}
$
For $n=k+1$
$
\begin{aligned}
\mathrm{P}(k+1) & =1.2+2.3+3.4+\ldots . .+k(k+1)+(k+1)(k+2) \\
& =\frac{k(k+1)(k+2)}{3}+(k+1)(k+2) \quad[\mathrm{Using}(i)] \\
& =(k+1)(k+2)\left[\frac{k}{3}+1\right]
\end{aligned}
$

$
=(k+1)(k+2)\left[\frac{k+3}{3}\right]=\frac{(k+1)(k+2)(k+3)}{3}
$
$\therefore \mathrm{P}(\mathrm{k}+1)$ is true
Thus $P(k)$ is true $\Rightarrow P(k+1)$ is true
Hence by principle of mathematical induction, $P(n)$ is true for all $n \in N$
Question 5.
Using the Mathematical induction, show that for any natural number $n \geq 2$,
$
\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \ldots\left(1-\frac{1}{n^2}\right)=\frac{n+1}{2 n}
$
Solution:
Let $\mathrm{P}(n)$ is the statement
Let $\mathrm{P}(n) \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\ldots+\frac{1}{1+2+3+\ldots+n}=\frac{n-1}{n+1}$
For $n=2$ Given $n \geq 2$
$
\begin{aligned}
& \text { LHS } \Rightarrow \mathrm{P}(2)=\frac{1}{1+2}=\frac{1}{3} \\
& \text { RHS } \Rightarrow \mathrm{P}(2)=\frac{2-1}{2+1}=\frac{1}{3}
\end{aligned}
$
$\mathrm{P}(n)$ is tru LHS $=$ RHS $\Rightarrow \mathrm{P}(n)$ is true for $n=2$
Assume $\mathrm{t}$ Assume that the given statement is true for $n=k$
(i.e.) $\left(1-\right.$ (i.e.) $\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}=\frac{k-1}{k+1}$ is true
To prove To prove $\mathrm{P}(k+1)$ is true
$
\mathrm{P}(k+1)=\mathrm{P}(k)+\left(t_{k+1}\right)
$

$
\begin{aligned}
& \left.=\frac{k-1}{k+1}+\frac{1}{1+2+\ldots+k+1}=\frac{k-1}{k+1}+\frac{1}{\frac{(k+1)(k+2)}{2}}\right] \\
=\frac{k+1}{2 k} ; & =\frac{k-1}{k+1}+\frac{2}{(k+1)(k+2)}=\frac{(k-1)(k+2)+2}{(k+1)(k+2)} \\
=\frac{k+1}{2 k} ; & =\frac{k^2-k+2 k-2+2}{(k+1)(k+2)}=\frac{k^2+k}{(k+1)(k+2)}
\end{aligned}
$
$\Rightarrow \mathrm{P}(\mathrm{k}+1)$ is true when $r(\mathrm{k})$ is true so by the principle of mathematical induction $r(\mathrm{n})$ is true.
Question 6.
Using the Mathematical induction, show that for any natural number $n \geq 2$,
$
\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\ldots+\frac{1}{1+2+3+\ldots+n}=\frac{n-1}{n+1} .
$
Solution:

Let $\mathrm{P}(n)$ is the statement
$
\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\ldots+\frac{1}{1+2+3+\ldots+n}=\frac{n-1}{n+1}
$
Given $n \geq 2$
$
\begin{aligned}
& \text { LHS } \Rightarrow \mathrm{P}(2)=\frac{1}{1+2}=\frac{1}{3} \\
& \text { RHS } \Rightarrow \mathrm{P}(2)=\frac{2-1}{2+1}=\frac{1}{3}
\end{aligned}
$
LHS $=$ RHS $\Rightarrow \mathrm{P}(n)$ is true for $n=2$
Assume that the given statement is true for $n=k$
(i.e.) $\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}=\frac{k-1}{k+1}$ is true
To prove $\mathrm{P}(k+1)$ is true
$
\begin{aligned}
\mathrm{P}(k+1) & =\mathrm{P}(k)+\left(t_{k+1}\right) \\
& =\frac{k-1}{k+1}+\frac{1}{1+2+\ldots+k+1}=\frac{k-1}{k+1}+\frac{1}{\frac{(k+1)(k+2)}{2}} \\
& =\frac{k-1}{k+1}+\frac{2}{(k+1)(k+2)}=\frac{(k-1)(k+2)+2}{(k+1)(k+2)} \\
& =\frac{k^2-k+2 k-2+2}{(k+1)(k+2)}=\frac{k^2+k}{(k+1)(k+2)}
\end{aligned}
$

$
=\frac{k(k+1)}{(k+1)(k+2)}=\frac{k}{k+1) 2}=\frac{k+1-1}{k+1+1}
$
$\Rightarrow \mathrm{P}(\mathrm{k}+1)$ is true when $\mathrm{P}(\mathrm{k})$ is true so by the principle of mathematical induction $\mathrm{P}(\mathrm{n})$ is true for $n \geq 2$.
Question 7.
Using the Mathematical induction, show that for any natural number $\mathrm{n}$
$
\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{n .(n+1) .(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}
$
Solution:

Let $\mathrm{P}(n)=\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{n \cdot(n+1) \cdot(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
For $n=1$
$
\begin{gathered}
P(1)=\frac{1}{1(1+1)(1+2)}=\frac{1(1+3)}{4(1+1)(1+2)} \\
\Rightarrow \frac{1}{1 \times 2 \times 3}=\frac{4}{4 \times 2 \times 3} \Rightarrow \frac{1}{6}=\frac{1}{6}
\end{gathered}
$
$\therefore \mathrm{P}(1)$ is true
Let $\mathrm{P}(n)$ be true for $n=k$
$
\therefore \mathrm{P}(k)=\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{k(k+1)(k+2)}+\frac{k(k+3)}{4(k+1)(k+2)}
$
For $n=k+1$
$
\begin{aligned}
& \text { RHS }=\frac{(k+1)(k+4)}{4(k+2)(k+3)} \\
& \text { LHS }=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \quad[\text { Using }(i)]
\end{aligned}
$

$
\begin{aligned}
& =\frac{1}{(k+1)(k+2)}\left[\frac{k^3+3 k}{4}+\frac{1}{k+3}\right] \\
& =\frac{1}{(k+1)(k+2)}\left[\frac{k^2+6 k^2+9 k+4}{4(k+3)}\right] \\
& =\frac{1}{(k+1)(k+2)}\left[\frac{(k+1)^2(k+4)}{4(k+3)}\right]=\left[\frac{(k+1)(k+4)}{4(k+2)(k+3)}\right]
\end{aligned}
$
$\therefore \mathrm{P}(\mathrm{k}+1)$ is true
Thus $\mathrm{p}(\mathrm{k})$ is true $\Rightarrow \mathrm{P}(\mathrm{k}+1)$ is true
Hence by principle of mathematical induction,
$\mathrm{p}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{z}$
Question 8.
Using the Mathematical induction, show that for any natural number $n$,
$
\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{6 n+4}
$
Solution:

Let $\mathrm{P}(n)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{6 n+4}$
For $n=1$
$
\begin{aligned}
& P(1)=\frac{1}{(3 \times 1-1)(3 \times 1+2)}=\frac{1}{(6 \times 1+4)} \\
& \Rightarrow \frac{1}{2 \times 5}=\frac{1}{10} \Rightarrow \frac{1}{10}=\frac{1}{10} \\
& \therefore \mathrm{P}(1) \text { is true } \\
& \therefore \quad \mathrm{P}(k)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 k-1)(3 k+2)} \\
& =\frac{k}{6 k+4} \\
&
\end{aligned}
$
For $n=k+1$
$
\begin{aligned}
& \mathrm{P}(k+1)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 k-1)(3 k+2)}+\frac{1}{[3(k+1)-1][3(k+1)+2]}=\frac{k+1}{6 k+10} \\
& =\frac{k}{6 k+4}+\frac{1}{(3 k+2)(3 k+5)}=\frac{1}{(3 k+2)}\left[\frac{k}{2}+\frac{1}{3 k+5}\right] \\
& =\frac{1}{(3 k+2)}\left[\frac{3 k^2+5 k+2}{2(3 k+5)}\right]=\frac{1}{(3 k+2)}\left[\frac{3 k^2+3 k+2 k+2}{2(3 k+5)}\right] \\
& =\frac{1}{(3 k+2)}\left[\frac{3 k(k+1)+2(k+1)}{2(3 k+5)}\right]=\frac{1}{(3 k+2)}\left[\frac{(k+1)(3 k+2)}{2(3 k+5)}\right] \\
& =\frac{k+1}{6 k+10}
\end{aligned}
$
$\therefore \mathrm{P}(\mathrm{k}+1)$ is true
Thus $\mathrm{P}(\mathrm{k})$ is true $\Rightarrow \mathrm{P}(\mathrm{k}+1)$ is true. Hence by principle of mathematical induction, $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in$ $\mathrm{N}$.

Question 9.
Prove by Mathematical Induction that
$
1 !+(2 \times 2 !)+(3 \times 3 !)+\ldots+(n \times n !)=(n+1) !-1
$
Solution:
$\mathrm{P}(\mathrm{n})$ is the statement
$
1 !+(2 \times 2 !)+(3 \times 3 !)+\ldots . .+(\mathrm{n} \times \mathrm{n} !)=(\mathrm{n}+1) !-1
$
To prove for $n=1$
$
\mathrm{LHS}=1 !=1
$
$
\begin{aligned}
& \text { RHS }=(1+1) !-1=2 !-1=2-1=1 \\
& \text { LHS }=\text { RHS } \Rightarrow \mathrm{P}(1) \text { is true }
\end{aligned}
$
Assume that the given statement is true for $\mathrm{n}=\mathrm{k}$
(i.e.) $1 !+(2 \times 2 !)+(3 \times 3$ ! $)+\ldots+(\mathrm{k} \times \mathrm{k} !)=(\mathrm{k}+1) !-1$ is true
To prove $\mathrm{P}(\mathrm{k}+1)$ is true
$
\mathrm{p}(\mathrm{k}+1)=\mathrm{p}(\mathrm{k})+\mathrm{t}_{(\mathrm{k}+1)}
$
$
\begin{aligned}
& \mathrm{P}(\mathrm{k}+1)=(\mathrm{k}+1) !-1+(\mathrm{k}+1) \times(\mathrm{k}+1) ! \\
& =(\mathrm{k}+1) !+(\mathrm{k}+1)(\mathrm{k}+1) !-1
\end{aligned}
$
$=(\mathrm{k}+1) ![1+\mathrm{k}+1]-1$
$=(\mathrm{k}+1) !(\mathrm{k}+2)-1$
$=(k+2) !-1$
$=(\mathrm{k}+1+1) !-1$
$\therefore \mathrm{P}(\mathrm{k}+1)$ is true $\Rightarrow \mathrm{P}(\mathrm{k})$ is true, So by the principle of mathematical induction $\mathrm{P}(\mathrm{n})$ is true.

Question 10.
Using the Mathematical induction, show that for any natural number $n, x^{2 n}-y^{2 n}$ is divisible by $x+y$.
Solution:
Let $P(n)=x^{2 n}-y^{2 n}$ is divisible by $(x+y)$
For $\mathrm{n}=1$
$P(1)=x^{2 \times 1}-y^{2 \times 1}$ is divisible by $(x+y)$
$\Rightarrow(x+y)(x-y)$ is divisible by $(x+y)$
$\therefore \mathrm{P}(1)$ is true
Let $P(n)$ be true for $n=k$
$\therefore \mathrm{P}(\mathrm{k})=\mathrm{x}^{2 \mathrm{k}}-\mathrm{y}^{2 \mathrm{k}}$ is divisible by $(\mathrm{x}+\mathrm{y})$
$\Rightarrow \mathrm{x}^{2 \mathrm{k}}-\mathrm{y}^{2 \mathrm{k}}=\lambda(\mathrm{x}+\mathrm{y})$
For $n=k+1$
$\Rightarrow P(k+1)=x^{2(k+1)}-y^{2(k+1)}$ is divisible by $(x+y)$
Now $\mathrm{x}^{2(\mathrm{k}+2)}-\mathrm{y}^{2(\mathrm{k}+2)}$
$=\mathrm{x}^{2 \mathrm{k}+2}-\mathrm{x}^{2 \mathrm{k}} \mathrm{y}^2+\mathrm{x}^{2 \mathrm{k}} \mathrm{y}^2-\mathrm{y}^{2 \mathrm{k}+2}$
$=\mathrm{x}^{2 \mathrm{k}} \cdot \mathrm{x}^2-\mathrm{x}^{2 \mathrm{k}} \mathrm{y}^2+\mathrm{x}^{2 \mathrm{k}} \mathrm{y}^2-\mathrm{y}^{2 \mathrm{k}} \mathrm{y}^2$
$=\mathrm{x}^{2 \mathrm{k}}\left(\mathrm{x}^2-\mathrm{y}^2\right)+\mathrm{y}^2 \lambda(\mathrm{x}+\mathrm{y})[\mathrm{U} \operatorname{sing}(\mathrm{i})]$
$\Rightarrow \mathrm{x}^{2 \mathrm{k}+2}-\mathrm{y}^{2 \mathrm{k}+2}$ is divisible by $(\mathrm{x}+\mathrm{y})$
$\therefore \mathrm{P}(\mathrm{k}+1)$ is true.
Thus $\mathrm{P}(\mathrm{k})$ is true $\Rightarrow P(k+1)$ is true. Hence by principle of mathematical induction, $P(n)$ is true for all $n \in$ $N$
Question 11.
By the principle of mathematical induction, prove that, for $n \geq 1$,
$
1^2+2^2+3^2+\ldots+n^2>\frac{n^3}{3}
$
Solution:

Let $\mathrm{P}(n)$ is the statement $1^2+2^2+3^2+\ldots+n^2>\frac{n^3}{3}$ To prove $P(1)$ is true
$
\mathrm{P}(1)=1^2=1>\frac{1^3}{3}\left(=\frac{1}{3}\right)
$
$1>\frac{1}{3}$ which is true
So $\mathrm{P}(1)$ is true
Assume that the given statement is true for $n=k$
(i.e.) $1^2+2^2+\ldots+k^2>\frac{k^3}{3}$ is true
To prove $\mathrm{P}(k+1)$ is true
$
\begin{aligned}
\mathrm{P}(k+1) & =\mathrm{P}(k)+(k+1) \\
\mathrm{P}(k+1) & =1^2+2^2+\ldots+k^2+(k+1)^2>\frac{k^3}{3}+(k+1)^2 \\
\text { RHS } & =\frac{k^3+3(k+1)^2}{3} \\
& =\frac{k^3+3\left(k^2+2 k+1\right)}{3}=\frac{k^3+3 k^2+6 k+3}{3} \\
& =\frac{k^3+3 k^2+3 k+3 k+1+2}{3}=\frac{k^3+3 k^2+3 k+1}{3}+\frac{3 k+2}{3}
\end{aligned}
$

$
\begin{aligned}
& =\frac{(k+1)^3}{3}+\frac{3 k+2}{3}>\frac{(k+1)^3}{3} \\
\Rightarrow \quad \mathrm{P}(k+1) & =1^2+2^2+\ldots+(k+1)^2>\frac{(k+1)^3}{3}
\end{aligned}
$
$\Rightarrow \mathrm{P}(k+1)$ is true whenever $\mathrm{P}(k)$ is true. So by the principle of mathematical inductions $\mathrm{P}(n)$ is true.
Question 12.
Use induction to prove that $n^3-7 n+3$, is divisible by 3 , for all natural numbers $n$.
Solution:
Let $P(n): n^3-7 n+3$
Step 1:
$\mathrm{P}(1)=(1)^3-7(1)+3$
$=1-7+3=-3$ which is divisible by 3
So, it is true for $\mathrm{P}(1)$.

Step 2:
$\mathrm{P}(\mathrm{k}): \mathrm{k}^3-7 \mathrm{k}+3=3 \lambda$. Let it be true
$
\Rightarrow \mathrm{k}^3=3 \lambda+7 \mathrm{k}-3
$
Step 3:
$
\begin{aligned}
& \mathrm{P}(\mathrm{k}+1)=(\mathrm{k}+1)^3-7(\mathrm{k}+1)+3 \\
& =\mathrm{k}^3+1+3 \mathrm{k}^2+3 \mathrm{k}-7 \mathrm{k}-7+3 \\
& =\mathrm{k}^3+3 \mathrm{k}^2-4 \mathrm{k}-3 \\
& \left.=(3 \lambda+7 \mathrm{k}-3)+3 \mathrm{k}^2-4 \mathrm{k}-3 \text { (from Step } 2\right) \\
& =3 \mathrm{k}^2+3 \mathrm{k}+3 \lambda-6 \\
& =3\left(\mathrm{k}^2+\mathrm{k}+\lambda-2\right) \text { which is divisible by } 3 .
\end{aligned}
$
So it is true for $\mathrm{P}(\mathrm{k}+1)$.
Hence, $P(k+1)$ is true whenever it is true for $P(k)$.
Question 13.
Use induction to prove that $5^{\mathrm{n}+1}+4 \times 6^{\mathrm{n}}$ when divided by 20 leaves a remainder 9 , for all natural numbers $n$.
Solution:
$P(\mathrm{n})$ is the statement $5^{\mathrm{n}+1}+4 \times 6^{\mathrm{n}}-9$ is $\div$ by 20
$
\mathrm{P}(1)=5^{1+1}+4 \times 6^1-9=5^2+24-9
$
$
=25+24-9=40 \div \text { by } 20
$
So $\mathrm{P}(1)$ is true
Assume that the given statement is true for $\mathrm{n}=\mathrm{k}$
(i.e) $5^{\mathrm{k}+1}+4 \times 6^{\mathrm{n}}-9$ is $\div$ by 20
$
\begin{aligned}
& \mathrm{P}(1)=5^{1+1}+4 \times 6^1-9 \\
& =25+24-9
\end{aligned}
$
So $\mathrm{P}(1)$ is true
To prove $P(k+1)$ is true
$
\begin{aligned}
& \mathrm{P}(\mathrm{k}+1)=5^{\mathrm{k}+1+1}+4 \times 6^{\mathrm{k}+1+1}-9 \\
& =5 \times 5^{\mathrm{k}+1}+4 \times 6 \times 6^{\mathrm{k}}-9
\end{aligned}
$

$
\begin{aligned}
& =100 \mathrm{C}+46^{\mathrm{k}}+36 \\
& =100 \mathrm{C}+4\left(9+6^{\mathrm{k}}\right) \\
& \text { Now for } \mathrm{k}=1 \Rightarrow 4\left(9+6^{\mathrm{k}}\right)=4(9+6) \\
& =4 \times 15=60 \div \text { by } 20 . \\
& \text { for } \mathrm{k}=2=4\left(9+6^2\right)=4 \times 45=180 \div 20
\end{aligned}
$
So by the principle of mathematical induction $4\left(9+6^{\mathrm{k}}\right)$ is $\div$ by 20
Now $100 \mathrm{C}$ is $\div$ by 20 .
So $100 \mathrm{C}+4\left(9+6^{\mathrm{k}}\right)$ is $\div$ by 20
$\Rightarrow P(k+1)$ is true whenever $P(k)$ is true. So by the principle of mathematical induction $P(n)$ is true.
Question 14
. Use induction to prove that $1 \theta+3 \times 4^{\mathrm{n}+2}+5$, is divisible by 9 , for all natural numbers $n$.

Solution:

$\mathrm{P}(\mathrm{n})$ is the statement $10^{\mathrm{n}}+3 \times 4^{\mathrm{n}+2}+5$ is $\div$ by 9
$
\begin{aligned}
& \mathrm{P}(1)=10^1+3 \times 4^2+5=10+3 \times 16+5 \\
& =10+48+5=63 \div \text { by } 9
\end{aligned}
$
So $\mathrm{P}(1)$ is true. Assume that $\mathrm{P}(\mathrm{k})$ is true
(i.e.) $10^k+3 \times 4^{k+2}+5$ is $\div$ by 9
(i.e.) $10^{\mathrm{k}}+3 \times 4^{\mathrm{k}+2}+5=9 \mathrm{C}$ (where $\mathrm{C}$ is an integer)
$
\Rightarrow 10^{\mathrm{k}}=9 \mathrm{C}-5-3 \times 4^{\mathrm{k}+2}
$
To prove $\mathrm{P}(\mathrm{k}+1)$ is true.
Now $\mathrm{P}(\mathrm{k}+1)=10^{\mathrm{k}+1}+3 \times 4^{\mathrm{k}+3}+5$
$=10 \times 10^{\mathrm{k}}+3 \times 4^{\mathrm{k}+2} \times 4+5$
$=10\left[9 \mathrm{C}-5-3 \times 4^{\mathrm{k}+2}\right]+3 \times 4^{\mathrm{k}+2} \times 4+5$
$=10\left[9 \mathrm{C}-5-3 \times 4^{\mathrm{k}+2}\right]+12 \times 4^{\mathrm{k}+2}+5$
$=90 \mathrm{C}-50-30 \times 4^{\mathrm{k}+2}+12 \times 4^{\mathrm{k}+2}+5$
$=90 \mathrm{C}-45-18 \times 4^{\mathrm{k}+2}$
$=9\left[10 \mathrm{C}-5-2 \times 4^{\mathrm{k}+2}\right]$ which is $\div$ by 9
So $\mathrm{P}(\mathrm{k}+1)$ is true whenever $\mathrm{P}(\mathrm{K})$ is true. So by the principle of mathematical induction $\mathrm{P}(\mathrm{n})$ is true.
Question 15.
Prove that using the Mathematical induction
$
\sin (\alpha)+\sin \left(\alpha+\frac{\pi}{6}\right)+\sin \left(\alpha+\frac{2 \pi}{6}\right)+\ldots+\sin \left(\alpha+\frac{(n-1) \pi}{6}\right)=\frac{\sin \left(\alpha+\frac{(n-1) \pi}{12}\right) \times \sin \left(\frac{n \pi}{12}\right.}{\sin \left(\frac{\pi}{12}\right)}
$
Solution:
$\mathrm{P}(n)$ is the statement

$\begin{aligned}
& \sin (\alpha)+\sin \left(\alpha+\frac{\pi}{6}\right)+\sin \left(\alpha+\frac{2 \pi}{6}\right)+\ldots+\sin \left(\alpha+\frac{(n-1) \pi}{6}\right)=\frac{\sin \left(\alpha+\frac{(n-1) \pi}{12}\right) \times \sin \left(\frac{n \pi}{12}\right)}{\sin \left(\frac{\pi}{12}\right)} \\
& \text { Put } n=1 \Rightarrow \mathrm{P}(1)=\sin \alpha=\text { LHS } \\
& \text { RHS }= \frac{\sin \left(\alpha+\frac{(1-1) \pi}{12}\right) \sin \frac{\pi}{12}}{\sin \frac{\pi}{12}}=\sin \alpha
\end{aligned}$

LHS $=\mathrm{RHS} \Rightarrow \mathrm{P}(1)$ is true
Assume that the statement is true for $n=k$
(i.e.) $\mathrm{P}(\mathrm{k})=\sin (\alpha)+\sin \left(\alpha+\frac{\pi}{6}\right)+\sin \left(\alpha+\frac{2 \pi}{6}\right)+\ldots+\sin \left(\alpha+\frac{(k-1) \pi}{6}\right)$ $=\frac{\sin \left(\alpha+\frac{(k-1) \pi}{12}\right) \times \sin \left(\frac{k \pi}{12}\right)}{\sin \left(\frac{\pi}{12}\right)}$ To prove $\mathrm{P}(k+1)$ is true is true
$
\begin{aligned}
& \operatorname{Now} \mathrm{P}(k+1)=\mathrm{P}(k)+t(k+1) \\
& \operatorname{Now} \mathrm{P}(k+1)=\mathrm{P}(k)+\sin \left(\alpha+\frac{k \pi}{6}\right) \\
&= \frac{\sin \left[\alpha+(k-1) \frac{\pi}{12}\right] \sin \frac{k \pi}{12}}{\sin \frac{\pi}{12}}+\sin \left(\alpha+\frac{k \pi}{6}\right) \\
&= \frac{\sin \left[\alpha+(k-1) \frac{\pi}{12}\right] \sin \frac{k \pi}{12}+\sin \left(\alpha+\frac{k \pi}{6}\right) \sin \frac{\pi}{12}}{\sin \frac{\pi}{12}}
\end{aligned}
$

$\begin{aligned}
\mathrm{Nr.}= & \frac{1}{2}\left[\cos \left(\alpha+(k-1) \frac{\pi}{12}-\frac{k \pi}{12}\right)-\cos \left(\alpha+\frac{(k-1) \pi}{12}+\frac{k \pi}{12}\right)\right. \\
& \left.+\cos \left(\alpha+\frac{k \pi}{6}-\frac{\pi}{12}\right)-\cos \left(\alpha+\frac{k \pi}{6}+\frac{\pi}{12}\right)\right] \\
= & \frac{1}{2}\left[\cos \left(\alpha+\frac{k \pi}{12}-\frac{\pi}{12}-\frac{k \pi}{12}\right)-\cos \left(\alpha+\frac{k \pi}{12}-\frac{\pi}{12}+\frac{k \pi}{12}\right)\right. \\
= & +\frac{1}{2}\left[\cos \left(\alpha-\frac{\pi}{12}\right)-\cos \left(\alpha+\frac{2 k \pi}{12}-\frac{\pi}{12}\right)\right. \\
& \left.\left.+\cos \left(\alpha+\frac{2 k \pi}{12}-\frac{\pi}{12}\right)-\cos \left(\alpha+\frac{2 k \pi}{12}+\frac{\pi}{12}\right)\right]-\cos \left(\alpha+\frac{2 k \pi}{12}+\frac{\pi}{12}\right)\right]
\end{aligned}$