How do you use the properties of summation to evaluate the sum of #Sigma i(i-1)^2# from i=1 to 15?

1 Answer
Nov 12, 2016

Answer:

# sum_(i=1)^15 i(i-1)^2 = 12040 #

Explanation:

We need to use these the standard results:
# sum_(r=1)^n r = 1/2n(n+1) #
# sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #
# sum_(r=1)^n r^3 = 1/4n^2(n+1)^2 #

We have;

# sum_(i=1)^n i(i-1)^2 = sum_(i=1)^n i(i^2-2i+1)#
# :. sum_(i=1)^n i(i-1)^2 = sum_(i=1)^n (i^3-2i^2+i)#
# :. sum_(i=1)^n i(i-1)^2 = sum_(i=1)^n i^3 - 2sum_(i=1)^n i^2 + sum_(i=1)^n i#

So for the sum requested, using the standard results;

# sum_(i=1)^15 i(i-1)^2 = 1/2(15)(15+1)-(2)(1/6)(15)(15+1)(30+1) + [1/2(15)(15+1)]^2 #

# :. sum_(i=1)^15 i(i-1)^2 = 1/2(15)(16)-(1/3)(15)(16)(31) + [1/2(15)(16)]^2 #
# :. sum_(i=1)^15 i(i-1)^2 = 120 - 2480 + 120^2 #
# :. sum_(i=1)^15 i(i-1)^2 = 120 - 2480 + 14400 #
# :. sum_(i=1)^15 i(i-1)^2 = 12040 #

Note:
We could equally derive a single formula the sum of #n# terms;

# sum_(i=1)^n i(i-1)^2 = 1/2n(n+1) -2 1/6n(n+1)(2n+1) + 1/4n^2(n+1)^2#
# :. sum_(i=1)^n i(i-1)^2 = 1/2n(n+1) - 1/3n(n+1)(2n+1) + 1/4n^2(n+1)^2#
# :. sum_(i=1)^n i(i-1)^2 = 1/12n(n+1) {6 - 4(2n+1) +3n(n+1) }#

# :. sum_(i=1)^n i(i-1)^2 = 1/12n(n+1) (6 - 8n-4 -3n^2 + 3n ) #
# :. sum_(i=1)^n i(i-1)^2 = 1/12n(n+1) (3n^2 - 5n + 2 ) #

And with #n=15# we get:

# sum_(i=1)^15 i(i-1)^2 = 1/12(15)(16) {3(15)(15) - 5(15) + 2 } #
# :. sum_(i=1)^15 i(i-1)^2 = (20) (675-75+2) #
# :. sum_(i=1)^15 i(i-1)^2 = (20) (602) #
# :. sum_(i=1)^15 i(i-1)^2 = 12040 #, as before