375 views

1 Answer

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

edited by

Related questions

3 votes
3 votes
1 answer
1
Devasish Ghosh asked Mar 4, 2017
517 views
Find all real solutions of the equation $x^{2} - |x-1| - 3 = 0$
0 votes
0 votes
1 answer
2
Tesla! asked Apr 24, 2018
1,483 views
The number of the ordered pair (X, Y), where X and Y are N*N real matrices such that XY-YX=​​​​​​​​​​​​​​ I isA) 0B) 1C) ND) Infinite
0 votes
0 votes
1 answer
3
Tesla! asked Apr 24, 2018
752 views
Let $(x_n)$ be a sequence of a real number such that the subsequence $(x_{2n})$ and $(x_{3n})$ converge to limit $K$ and $L$ respectively. Then$(x_n)$ always convergeIf $...
1 votes
1 votes
1 answer
4