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

Question

Charles Arocha · Accepted Answer

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

Axel Alvarez · Answer

undefined To evaluate the sum of Σi(i^2 + 1) from i = 1 to 10, you can use the properties of summation. First, expand the expression inside the sigma notation. Then, apply the distributive property of summation to split the expression into two separate sums. Next, evaluate each sum separately using the formulas for the sum of consecutive integers and the sum of consecutive squares. Finally, sum up the results of the two separate sums to find the overall sum of the expression.