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If n ge 1 is a positive integer, then prove that 3^(n) ge 2^(n) + n . 6^((n - 1)/(2)) |
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Answer» Solution :We KNOW that , `a^(n) - B^(n) = (a-b) (a^(n-1) + a^(n-2) b + a^(n-3) b^(2) + .. . + b^(n-1))` `THEREFORE 3^(n) - 2^(n)= 3^(n-1) + 3^(n-2) 2 + 3^(n-3) 2^(2) + … + 2^(n-1)` Using `A.M. ge G.M` , we GET `(3^(n-1)+ 3^(n-2) . 2 + … + 2^(n-1))/(n) ge [(3 * 3^(2) * ... * 3^(n-1)) (2* 2^(2) * ... * 2^(n-1))]^(1//n)` `3^((n-1)/(2)) * 2((n-1)/(2)) = 6^((n-1)/(2))` `implies 3^(n) ge 2^(n) + n* 6 ((n-1)/(2))`. |
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