0 votes 0 votes Is there any shortcut to compute this or does it have to get the sum by calculating each values? Others engineering-mathematics + – Abhijit Sen asked Apr 2, 2017 Abhijit Sen 375 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes a1=1 a2=6 , a3=12 an=n*(n+1) $\sum_{1}^{n} n*(n+1)$ $\sum_{1}^{n} n + \sum_{1}^{n} n^{2}$ $\sum_{1}^{n} n = \frac{n*(n+1)}{2}$ & $\sum_{1}^{n} n^{2} = \frac{n*(n+1)(2n+1)}{6}$ $\frac{20*21}{2}+\frac{20*21*41}{6}$ $210+2870$ = 3080 Correct Option C Tesla! answered Apr 2, 2017 edited Apr 6, 2017 by Tesla! Tesla! comment Share Follow See all 0 reply Please log in or register to add a comment.