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Question

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


A
2
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B
4
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C
8
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D
12
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Solution

The correct option is A 2
$$3^{2n} =(1+2)^{2n} =  ^{2n}C_0+^{2n}C_12+^{2n}C_22^2+........+ ^{2n}C_{2n}2^{2n}$$
$$\quad \quad =  1+4n+^{2n}C_22^2+........+ ^{2n}C_{2n}2^{2n}$$
$$\therefore 3^{2n}-2n+1 =2+2n+^{2n}C_22^2+........+ ^{2n}C_{2n}2^{2n}$$
$$\quad \quad =2(1+n+^{2n}C_22+........+ ^{2n}C_{2n}2^{2n-1}) =2\times $$ some integer
Hence for all natural number $$3^{2n}-2n+1$$ is always divisible by $$2$$

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