Question

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

A
2
B
4
C
8
D
12

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 integerHence for all natural number $$3^{2n}-2n+1$$ is always divisible by $$2$$Maths

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