How do you use the properties of summation to evaluate the sum of #Sigma (i^3-2i)# from i=1 to 15?

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Answer 1 · 2016-12-24T09:42:17

The answer is #=14160#

We need

#sum_1^nr^3=1/4n^2(n+1)^2#

#sum_1^nr=1/2n(n+1)#

Therefore,

#sum_1^15(i^3-2i)#

#=sum_1^15i^3-2sum_1^15i#

#=1/4*15^2(16^2)-2*1/2*15*16#

#=15*16(15*16/4-1)#

#=15*16(60-1)#

#=15*16*59#

#=14160#