The expression {10^n} + 3 times {4^{n + 2}} + | vedclass

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(C) Let $P(n) = {10^n} + 3 	imes {4^{n + 2}} + 5$.For $n = 1$,$P(1) = {10^1} + 3 	imes {4^3} + 5 = 10 + 3 	imes 64 + 5 = 10 + 192 + 5 = 207$.Since

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$9$

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(C) Let $P(n) = {10^n} + 3 	imes {4^{n + 2}} + 5$.For $n = 1$,$P(1) = {10^1} + 3 	imes {4^3} + 5 = 10 + 3 	imes 64 + 5 = 10 + 192 + 5 = 207$.Since $207 = 9 	imes 23$,it is divisible by $9$.For $n = 2$,$P(2) = {10^2} + 3 	imes {4^4} + 5 = 100 + 3 	imes 256 + 5 = 100 + 768 + 5 = 873$.Since $873 = 9 	imes 97$,it is divisible by $9$.Thus,the expression is divisible by $9$ for all $n \in N$.

The expression ${10^n} + 3 \times {4^{n + 2}} + 5$ is divisible by which of the following for all $n \in N$?

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