Question

How do you find the partial sum of `Sigma (2n+5)` from n=1 to 20?

Answer 1

`(20/2)(7 + 45) =520`

Explanation:

The partial sum formula for n terms looks like this:
`sum_n = n/2(a_1 + a_n)` In words, it's the number of terms times the average of the first and last term.

In your case, with n=20, you need to find what `2n+5` is for `n=1` and `n=20`.
When `n=1, 2n+5=7.` Likewise, when `n=20, 2n+5=45.`

Your formula ends up looking like this: `(20/2)(7 + 45) =520`.

Answer 2

`sum_1^20 (2n+5)=520`

Explanation:

Before we proceed a few identities

`sum_1^n n^0=sum_1^n 1=n`

`sum_1^n n=(n(n+1))/2`

`sum_1^n n^2=(n(n+1)(2n+1))/6`

`sum_1^n n^3=((n(n+1))/2)^2`

`sum_1^n n^4=(n(n+1)(2n+1)(3n^2+3n-1))/6`

Hence, `sum_1^n (2n+5)`

= `2sum_1^n n+5xxSigma_1^n 1`

= `2xx(n(n+1))/2+5n`

= `n^2+n+5n`

= `n^2+6n`

and `sum_1^20 (2n+5)=20^2+6xx20=520`