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Mathematics Question on Sum of First n Terms of an AP

The sum 1+13+231+2+13+23+331+2+3+....+13+23+33+....+1531+2+3+...+15−12(1+2+3+...+15)1+ \frac{1^{3} +2^{3}}{1+2} + \frac{1^{3}+2^{3}+3^{3}}{1+2+3} +.... + \frac{1^{3} +2^{3}+3^{3} +....+15^{3}}{1+2+3+...+15} - \frac{1}{2} \left(1+2+3+...+15\right)

A

1240

B

1860

C

660

D

620

Answer

620

Explanation

Solution

Sum =∑n=11513+23+...n31+2+...+n−12.15.162= \sum^{15}_{n=1} \frac{1^{3} + 2^{3} +...n^{3}}{1+2+...+n} - \frac{1}{2}. \frac{15.16}{2} =∑n=115n(n+1)2−60=∑n=115n(n+1)(n+2−(n−1))6−60=\sum^{15}_{n=1} \frac{n\left(n+1\right)}{2} -60 =\sum^{15}_{n=1} \frac{n\left(n+1\right)\left(n+2-\left(n-1\right)\right)}{6} -60 =15.16.176−60=620= \frac{15.16.17}{6} - 60 = 620