Question

Use summation formulas to rewrite the expression without the summation notation .

sum i=1 ^ n 2i^ 3 -3i n^ 4

Answer 1

Factor out 1/n⁴, and distribute the sum over each term:

`\displaystyle\sum_(i=1)^n(2i^3-3i)/(n^4)=\frac1{n^4}\sum_(i=1)^n(2i^3-3i)`

`\displaystyle\sum_(i=1)^n(2i^3-3i)/(n^4)=\frac2{n^4}\sum_(i=1)^ni^3-\frac3{n^4}\sum_(i=1)^ni`

Recall the following formulas:

`\displaystyle\sum_(i=1)^ni=\frac{n(n+1)}2`

`\displaystyle\sum_(i=1)^ni^3=\frac{n^2(n+1)^2}4`

So we have

`\displaystyle\sum_(i=1)^n(2i^3-3i)/(n^4)=(n^2(n+1)^2)/(2n^4)-(3n(n+1))/(2n^4)=\boxed{((n+1)(n^2+n-3))/(2n^3)}`