Question

Prove that for each positive integer n, the integer 32n+1 + 2n+2 is divisible by 7.​

Answer 1

To prove:

For every positive integer n, the integer $\sf{3^{(2n+1)} + 2^{(n+2)}}$ will be divisible by 7

Proof:

Considering the method of mathematical induction

Let us check for n = 1

→ $\sf{3^{(2(1)+1)} + 2^{((1)+2)}}$ = $\sf{3^3 + 2^3}$ = 35  (divisible by 7)

Checking for n = 2

→ $\sf{3^{(2(2)+1)} + 2^{((2)+2)}}$ = $\sf{3^5 + 2^4}$ = 243 + 16 = 259   (divisible by 7)

Now, let us assume that

$\sf{3^{(2x+1)} + 2^{(x+2)}}$ is divisible by 7 for some positive integer 'x' such that,

$\sf{3^{(2x+1)} + 2^{(x+2)}=7k}$  for any integer k.

that is, $\sf{ 2^{(x+2)} = 7k - 3^{(2x+1)} \;\;...eqn(1)}$

Now, We need to show that integer will be divisible by 7 for n = x + 1 also, i.e., $\sf{3^{(2(x+1)+1)} + 2^{((x+1)+2)}}$ is also divisible by 7

so,

$\implies \sf{3^{(2(x+1)+1)} + 2^{((x+1)+2)}}$

$\qquad \qquad = \sf{3^{(2x+3)} + 2^{(x+3)}}$

$\qquad \qquad = \sf{3^{2} \;.\; 3^{(2x+1)} + 2^{(x+2)} \;.\; 2}$

$\qquad \qquad = \sf{3^{2} \;.\; 3^{(2x+1)} + 2^{(x+2)} \;.\; 2}$

using equation (1)

$\qquad \qquad = \sf{3^{2} \;.\; 3^{(2x+1)} + (7k - 3^{(2x+1)} )\;.\; 2}$

$\qquad \qquad = \sf{9 \;.\; 3^{(2x+1)} + 14k - 2 (3^{(2x+1)}) }$

$\qquad \qquad = \sf{ (9-2) \;.\; 3^{(2x+1)}+ 14k }$

$\qquad \qquad = \sf{ 7 \;.\; 3^{(2x+1)}+ 14k }$

$\qquad \qquad = \sf{ 7\; ( 3^{(2x+1)}+ 2k) }$

That is the multiple of 7.

Therefore, By the concept of mathematical induction, $\sf{3^{(2n+1)} + 2^{(n+2)}}$ is divisible by 7 for all positive integer n.