Use the identities to evaluate the sums below: sum_i=1^n(8i^3+2/(n^4)*sum_i=1^n(12i^2-10/(n^3))   sum_i=1^n (1=n)sum_i=1^n (i=n(n+1)/2)sum_i=1^n (i^2=n(n+1)(2n+1)/6)sum_i=1^n (i^3=n^2(n+1)^2/(4)

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The identities are,

`sum_(i=1)^n 1=n`
`sum_(i=1)^n i=(n(n+1))/2`
`sum_(i=1)^n i^2=(n(n+1)(2n+1))/6`
`sum_(i=1)^n i^3=(n^2(n+1)^2)/4`

 

now your question is,

`sum_(i=1)^n(8i^3+2/(n^4))*sum_(i=1)^n(12i^2-10/(n^3))`

`=(sum_(i=1)^n8i^3 + sum_(i=4)^n2/n^4)*(sum_(i=1)^n12i^2 - sum_(i=1)^n10/n^3)`

`=(8sum_(i=1)^ni^3 + 2/(n^4)sum_(i=4)^n1)*(12sum_(i=1)^ni^2 - 10/(n^3)sum_(i=1)^n)`

`=(8(n^2(n+1)^2)/4 + 2/(n^4)*n)*(12(n(n+1)(2n+1))/6-10/(n^3)*n)`

`=((n^2(n+1)^2)/2 + 2/(n^3))*((n(n+1)(2n+1))/2-10/(n^2))`

`=((n^5(n+1)^2 + 4)/(2n^3))*((n^3(n+1)(2n+1)-20)/(2n^2))`

`=((n^5(n+1)^2 + 4)*(n^3(n+1)(2n+1)-20))/(4n^5)`

 

 

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