Question

How do you evaluate `\sum _ { n = 1} ^ { 99} ( 6n + 4)`?

Answer 1

`sum_(n=1)^99(6n+4)=30096`

Explanation:

`sum_(n=1)^99(6n+4)= 6*sum_(n=1)^99n+sum_(n=1)^99 4`

`=6*(n(n+1))/2+4n=3 *99*100+396`

`=6*(n(n+1))/2+4n=29700+396=30096`

`:. sum_(n=1)^99(6n+4)=30096` [Ans]