Prove that 11n+2 + 122n+1 is divisible by 133 for all n belongs to N - Maths - Principle of Mathematical Induction

Question

Prove that 11n+2 + 122n+1 is divisible by
133 for all n belongs to N

Duke Biswass · Accepted Answer

$11^{n + 2} + 12^{2 n + 1}$

for n =1 we get $11^{2} + 12^{3} = 1331 + 1728 = 3059$

3059/133 = 23

hence it is true for n =1

let us assume that this expression is true for n = k

$11^{k + 2} + 12^{2 k + 1}$ is divisible by 133

for n = k+1

$11^{k + 3} + 12^{2 k + 3}$

= $(11^{k} \times 1331) + (12^{2 k} \times 1728)$

= $(11^{k} \times 1331) + ({(11 + 1)}^{2 k} \times 1728)$

which is divisible by 1331+1728 = 3059 which is divisible by 133

hence our assumption is correct and $11^{n + 2} + 12^{2 n + 1}$ is divisible by 133

Prove that 11n+2 + 122n+1 is divisible by 133 for all n belongs to N .