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

2 Answers

Answer:

#(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:

#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#