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

2 Answers

(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.

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