If the equation 1 3^2 + 2 5^2 + 3 7^2 + + n(2n+1)^2 1^2 3 + 2^2 5 + 3^2 7 + + n^2(2n+1) = 83 39 holds true for some positive integer n , then the value of n is

Step-by-step solution

The general term of the series in the numerator is T_k = k(2k+1)^2 = 4k^3 + 4k^2 + k . The general term of the series in the denominator is T'_k = k^2(2k+1) = 2k^3 + k^2 . The sum of the numerator up to n terms is: _ k=1 ^n (4k^3 + 4k^2 + k) = 4 [ n(n+1) 2 ]^2 + 4 [ n(n+1)(2n+1) 6 ] + n(n+1) 2 = n^2(n+1)^2 + 2 3 n(n+1)(2n+1) + n(n+1) 2 = n(n+1) 6 [ 6n(n+1) + 4(2n+1) + 3 ] = n(n+1) 6 (6n^2 + 14n + 7) The sum of the denominator up to n terms is: _ k=1 ^n (2k^3 + k^2) = 2 [ n(n+1) 2 ]^2 + n(n+1)(2n+1) 6 = n^2(n+1)^2 2