Consider the series 1^3 + 2^2 + 3^3 + 4^2 + 5^3 + 6^2 + Let S_ 2n denote the sum of the first 2n terms of this series. If S_ 2n can be expressed in the form n 3 (an^3 + bn^2 + cn + d) , where a, b, c, and d are integers, then the value of a+b+c+d is equal to

Step-by-step solution

The given series consists of two interleaved sub-series: the odd-positioned terms are cubes of odd integers, and the even-positioned terms are squares of even integers. The sum of the first 2n terms can be written as: S_ 2n = _ r=1 ^ n [ (2r-1)^3 + (2r)^2 ] Expanding the terms inside the summation: (2r-1)^3 = 8r^3 - 12r^2 + 6r - 1 (2r)^2 = 4r^2 Adding them together gives the general term for the pairs: (2r-1)^3 + (2r)^2 = 8r^3 - 8r^2 + 6r - 1 Now, sum this expression from r=1 to n : S_ 2n = 8 _ r=1 ^ n r^3 - 8 _ r=