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Show that 9^n + 1 - 8n - 9 is divisible by 64, whenever n is a positive integer.

Q13.    Show that 9^{n+1} - 8n - 9is divisible by 64, whenever n is a positive integer.

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N neha

If we want to prove that  9^{n+1} - 8n - 9is divisible by 64, then we have to prove that  9^{n+1} - 8n - 9=64k

As we know, from binomial theorem, 

(1+x)^m=^mC_0+^mC_1x+^mC_2x^2+^mC_3x^3+....^mC_mx^m

Here putting x = 8 and replacing m by n+1, we get,

9^{n+1}=^{n+1}C_0+\:^{n+1}C_18+^{n+1}C_28^2+.......+^{n+1}C_{n+1}8^{n+1}

9^{n+1}=1+8(n+1)+8^2(^{n+1}C_2+\:^{n+1}C_38+^{n+1}C_48^2+.......+^{n+1}C_{n+1}8^{n-1})

9^{n+1}=1+8n+8+64(k)

Now, Using This,

9^{n+1} - 8n - 9=9+8n+64k-9-8n=64k

Hence

9^{n+1} - 8n - 9 is divisible by 64.

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