Question

Using principle of mathematical induction, $4^n+15n-1$ is divisible by

• A. $9\forall n\in N$
• B. $11\forall n\in N$
• C. $13\forall n\in N$
• D. $15\forall n\in N$

Answer 1

Answer: (A)

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, $4^{k+1}+15\left(k+1\right)-1$
$=4^k\cdot 4+15\left(k+1\right)-1$
$=\left(9\lambda -15k+1\right)\times 4+15\left(k+1\right)-1$
$=36\lambda -45k+18$
$=9\left(4\lambda -5k+2\right)$, 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 $n\in N$.