If n in N,then {11^{n + 2}} + {12^{2n + 1}} i | vedclass

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(C) Let $f(n) = 11^{n+2} + 12^{2n+1}$.For $n = 1$,$f(1) = 11^{1+2} + 12^{2(1)+1} = 11^3 + 12^3 = 1331 + 1728 = 3059$.Dividing $3059$ by $133$,we get $

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$133$

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(C) Let $f(n) = 11^{n+2} + 12^{2n+1}$.For $n = 1$,$f(1) = 11^{1+2} + 12^{2(1)+1} = 11^3 + 12^3 = 1331 + 1728 = 3059$.Dividing $3059$ by $133$,we get $3059 \div 133 = 23$.Since $3059$ is divisible by $133$,the expression is divisible by $133$ for $n=1$.For $n=2$,$f(2) = 11^4 + 12^5 = 14641 + 248832 = 263473$.Dividing $263473$ by $133$,we get $263473 \div 133 = 1981$.Thus,the expression is divisible by $133$ for all $n \in N$.

If $n \in N$,then ${11^{n + 2}} + {12^{2n + 1}}$ is divisible by

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