The value of _ i=1 ¹⁵ _ j=1 ^ i j(i-j) is equal to :

Step-by-step solution

First, we evaluate the inner sum with respect to j , treating i as a constant: _ j=1 ^ i (ij - j^2) = i _ j=1 ^ i j - _ j=1 ^ i j^2 Using the standard summation formulas: = i ( i(i+1) 2 ) - i(i+1)(2i+1) 6 Factoring out the common terms: = i(i+1) 2 ( i - 2i+1 3 ) = i(i+1) 2 ( 3i - 2i - 1 3 ) = i(i+1)(i-1) 6 = i(i^2 - 1) 6 = i^3 - i 6 Now, we substitute this back into the outer sum: _ i=1 ¹⁵ i^3 - i 6 = 1 6 ( _ i=1 ¹⁵ i^3 - _ i=1 ¹⁵ i ) Using the formulas for the sum of cubes and the sum of the first n natural number