For all positive integral values of n, $3^{2 n} - 2 n + 1$ is divisible by

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Solution

The correct option is A 2
$3^{2 n} = (1 + 2)^{2 n} =^{2 n} C_{0} +^{2 n} C_{1} 2 +^{2 n} C_{2} 2^{2} + . . . . . . . . +^{2 n} C_{2 n} 2^{2 n}$
$= 1 + 4 n +^{2 n} C_{2} 2^{2} + . . . . . . . . +^{2 n} C_{2 n} 2^{2 n}$
$∴ 3^{2 n} - 2 n + 1 = 2 + 2 n +^{2 n} C_{2} 2^{2} + . . . . . . . . +^{2 n} C_{2 n} 2^{2 n}$
$= 2 (1 + n +^{2 n} C_{2} 2 + . . . . . . . . +^{2 n} C_{2 n} 2^{2 n - 1}) = 2 \times$ some integer
Hence for all natural number $3^{2 n} - 2 n + 1$ is always divisible by $2$

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