Question

Power series help?

n
SUM 3k(k^2+4),
k=1

give answer in form 3/4n(n+1)(An^2+Bn+C)
k=1

Answer 1

See below.

Explanation:

If `S_n=sum_(k=1)^n 3k(k^2+4) = 3/4n(n+1)(A n^2+B n + C)`

then

`S_1=3/4(1)(2)(A+B+C) = 15`
`S_2=3/4(2)(3)(4A+2B+C)=63`
`S_3 =3/4(3)(4)(9A+3B+C)=180`

now solving

`{(3(A+B+C)=30),(9(4A+2B+C)=126),(9(9A+3B+C)=180):}`

for `A,B,C` we get at

`A = 1, B=1, C=8` hence the formula reads

`sum_(k=1)^n 3k(k^2+4) = 3/4n(n+1)(n^2+n + 8)`