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Q.
Using principle of mathematical induction, $4^{n}+15n-1$ is divisible by
Principle of Mathematical Induction
Solution:
Let $P\left(n\right)$ be the statement given by
$P\left(n\right) : 4^{n} +15n -1$ is divisible by $9$
For $n = 1, P\left(1\right): 4^{1} + 15 \times 1 - 1 = 18$, which is divisible by $9$
$\therefore P\left(1\right)$ is true .
Let $P\left(k\right)$ be true Then,
$4^{k} + 15k- 1 $ is divisible by $9$
$\Rightarrow 4^{k}+ 15k-1 = 9\lambda$, for some $\lambda\in N $
We shall now show that $P\left(k + 1\right)$ is true, for this we have to show that
$4^{k+1}+15 \left(k+1\right)-1$ is divisible by $9$.
Now, $4^{k+1} +15\left(k+1\right)-1$
$ = 4^{k}\cdot4 +15\left(k+1\right)-1$
$ = \left(9\lambda-15k +1\right) \times 4+15\left(k+1\right) -1$
$= 36\lambda -45k+18$
$=9\left(4\lambda-5k+2\right)$, which is divisible by $9$
$ \therefore P\left(k+1\right)$ is true.
Thus, $P\left(k\right)$ is true $\Rightarrow P\left(k + 1\right)$ is true.
Hence, by the principle of mathematical induction $P\left(n\right)$ is true for all $n\in N$.