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

1 Answer
Steve M
Oct 31, 2016

# sum_(i=1)^10(i^2-1) = 375 #

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^2# we have:

# sum_(i=1)^n(i^2-1) = sum_(i=1)^n i^2 - sum_(i=1)^n 1 #
# :. sum_(i=1)^n(i^2-1) = 1/6n(n+1)(2n+1) - n #
# :. sum_(i=1)^n(i^2-1) = 1/6n{(n+1)(2n+1) - 6} #
# :. sum_(i=1)^n(i^2-1) = 1/6n{2n^2+3n+1 - 6} #
# :. sum_(i=1)^n(i^2-1) = 1/6n{2n^2+3n- 5} #
# :. sum_(i=1)^n(i^2-1) = 1/6n(n-1)(2n+5) #

So with n=10 we have:
# sum_(i=1)^10(i^2-1) = 1/6(10)(9)(25) #
# sum_(i=1)^10(i^2-1) = 375 #

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