The sum 1^3 1 + 1^3 + 2^3 1 + 3 + 1^3 + 2^3 + 3^3 1 + 3 + 5 + up to 8 terms, is:

Step-by-step solution

The n -th term of the given series is: T_n = 1^3 + 2^3 + + n^3 1 + 3 + 5 + + (2n - 1) Using the formulas for the sum of cubes of first n natural numbers and the sum of first n odd numbers: T_n = ( n(n+1) 2 )^2 n^2 = n^2(n+1)^2 4n^2 = (n+1)^2 4 The sum of the series up to 8 terms is: S₈ = _ n=1 ⁸ T_n = _ n=1 ⁸ (n+1)^2 4 S₈ = 1 4 ( 2^2 + 3^2 + + 9^2 ) S₈ = 1 4 ( _ k=1 ⁹ k^2 - 1^2 ) Using the formula _ k=1 ^ n k^2 = n(n+1)(2n+1) 6 : S₈ = 1 4 ( 9 10 19 6 - 1 ) S₈ = 1 4 (285 - 1) = 284 4 = 71 Answer: 71