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

Question

Cameron Alexander · Accepted Answer

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

Samantha Adkison · Answer

undefined To evaluate the sum of Σ(i^2 - 1) from i=1 to 10, we first find the individual terms of the series by plugging in values of i from 1 to 10:

When i = 1: (1^2 - 1) = 0
When i = 2: (2^2 - 1) = 3
When i = 3: (3^2 - 1) = 8
When i = 4: (4^2 - 1) = 15
When i = 5: (5^2 - 1) = 24
When i = 6: (6^2 - 1) = 35
When i = 7: (7^2 - 1) = 48
When i = 8: (8^2 - 1) = 63
When i = 9: (9^2 - 1) = 80
When i = 10: (10^2 - 1) = 99

Next, we sum up these individual terms:

0 + 3 + 8 + 15 + 24 + 35 + 48 + 63 + 80 + 99 = 375

So, the sum of Σ(i^2 - 1) from i=1 to 10 is 375.