For every natural number $n (> 1)$ , prove that $4^{n} + 15 n - 1$ is divisible by $9$ .

Open in App

Solution

Proving by induction,
for $n = 1$ , $4^{n} + 15 n - 1 = 18$
Let for $n = k$ , $4^{k} + 15 k - 1 = 9 λ_{1}$
for $n = k H = 4^{k + 1} + 15 (k + 1) - 1$
$= {4.4}^{k} + 15 k + 14$
$= 4 (9 λ_{1} + 1 - 15 k) + 15 k + 14$
$= 36 λ_{1} + 4 - 60 k + 15 k + 14 = 36 λ_{1} - 45 k + 18$
$= 9 (4 λ_{1} - 5 k + 2) = 9 λ_{1}$ .

Suggest Corrections

Similar questions

Prove by using the principle of mathematical induction $\forall n \in N$

$2 + 5 + 8 + 11 + . . . + (3 n - 1) = \frac{1}{2} n (3 n + 1)$

Using principle of mathematical induction, prove that $4^{n} + 15 n - 1$ is divisible by 9 for all natural numbers n.

Prove by using the principle of mathematical induction $\forall n \in N$

$2 + 5 + 8 + 11 + . . . + (3 n - 1) = \frac{1}{2} n (3 n + 1)$

Using principle of mathematical induction, prove that $4^{n} + 15 n - 1$ is divisible by 9 for all natural numbers n.

2^{3 n} - 1

7

n

3^{2 n + 2} - 8 n - 9

4^{n} + 15 n - 1

9

n

Join BYJU'S Learning Program