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Show that `2^(4n+4)-15n-16, ` where n `in` N is divisible by 225. |
Answer» We have `2^(4n+4)-15n - 16 = 2^(4(n+1))-15n - 16 = 16^((n+1))- 15n -16.` Now, `16^(n+1)-15n-16=(1+15)^(n+1)-15n -16` `=^(n+1) C_(0)+^(n+1) C_(1) xx 15 + ^(n+1)C_(2) xx (15)^(2) + ^(n+1)C_(3) xx(15)^(3)+...+.^(n+1)C_(n+1) xx(15)^(n+1)-15n-16` `= 1+(n+1) xx 15 + .^(n+1)C_(2) xx(15)^(2)+ ^(n+1) C_(3) xx(15)^(3)+...+ (15)^(n+1)-15n-16` `=.^(n+1)C_(2) xx (15)^(2) = .^(n+1)C_(3) xx (15)^(3) +...+(15)^(n+1)` `(15)^(2) xx[.^(n+1)C_(2) + .^(n+1)C_(3) xx 15 +...+(15)^(n-1) ]` `=225 xx` (an integar). Hence, `(2^(4n+4)-15n-16)` is divisible by 225. |
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