If n in N,then {7^{2n}} + {2^{3n - 3}} cdot { | vedclass

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(A) Let $f(n) = 7^{2n} + 2^{3n-3} \cdot 3^{n-1}$.For $n = 1$,$f(1) = 7^{2(1)} + 2^{3(1)-3} \cdot 3^{1-1} = 7^2 + 2^0 \cdot 3^0 = 49 + 1 = 50$.For $n =

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$25$

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(A) Let $f(n) = 7^{2n} + 2^{3n-3} \cdot 3^{n-1}$.For $n = 1$,$f(1) = 7^{2(1)} + 2^{3(1)-3} \cdot 3^{1-1} = 7^2 + 2^0 \cdot 3^0 = 49 + 1 = 50$.For $n = 2$,$f(2) = 7^{2(2)} + 2^{3(2)-3} \cdot 3^{2-1} = 7^4 + 2^3 \cdot 3^1 = 2401 + 8 \cdot 3 = 2401 + 24 = 2425$.Since $50$ and $2425$ are both divisible by $25$,the expression is always divisible by $25$.

If $n \in N$,then ${7^{2n}} + {2^{3n - 3}} \cdot {3^{n - 1}}$ is always divisible by

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