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

Question

Christopher Agresta · Accepted Answer

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

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#

Ezekiel Alexander · Answer

#(20/2)(7 + 45) =520# 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#.

Ezra Adams · Answer

undefined To find the partial sum of the series Σ(2n + 5) from n = 1 to 20, you can use the formula for the sum of an arithmetic series:

S = n/2 * (first term + last term)

First, find the first term of the series when n = 1:

2(1) + 5 = 7

Next, find the last term of the series when n = 20:

2(20) + 5 = 45

Now, substitute these values into the formula:

S = 20/2 * (7 + 45)
S = 10 * 52
S = 520