How do you use the properties of summation to evaluate the sum of #Sigma i(i^2+1)# from i=1 to 10?

1 Answer
Oct 31, 2016

Answer:

# sum_(i=1)^10i(i^2+1) = 3080 #

Explanation:

We could just add up the 10 terms but the numbers get a bit horrific, so it is actually easier in this case to derive a general formula standard formula for #sumi# and #sumi^3# we have:

# sum_(i=1)^ni(i^2+1) = sum_(i=1)^n (i^3+i) #
# :. sum_(i=1)^ni(i^2+1) = sum_(i=1)^n i^3 + sum_(i=1)^n i #
# :. sum_(i=1)^ni(i^2+1) = (n^2(n+1)^2)/4 + (n(n+1))/2 #
# :. sum_(i=1)^ni(i^2+1) = (n(n+1))/4 { n(n+1) + 2 } #
# :. sum_(i=1)^ni(i^2+1) = (n(n+1))/4 ( n^2 + n + 2) #

So with n=10 we have:
# :. sum_(i=1)^10i(i^2+1) = (10(11))/4 ( 100 + 10 + 2) #
# :. sum_(i=1)^10i(i^2+1) = (10(11)(112))/4 #
# :. sum_(i=1)^10i(i^2+1) = 3080 #