Answer:
S(n)=(25/3)n(n+1)(2n+19)
Step-by-step explanation:
sum_(n=1)^k n (6 + n) = 1/6 k (k + 1) (2 k + 19)
[tex]sum_(n=1)^k n (6 + n) = (1/6) k (k + 1) (2 k + 19)\\50(1/6) k (k + 1) (2 k + 19)\\=(25/3)k (k + 1) (2 k + 19)[/tex]
This is the equation in general:
substitute n for sum from 1 to n
For sum from k to n, substitute:
S(n)-S(k-1)
