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:

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

$\displaystyle\sum_{i=1}^n\frac{2i^3-3i}{n^4}=\frac2{n^4}\sum_{i=1}^ni^3-\frac3{n^4}\sum_{i=1}^ni$

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

$\displaystyle\sum_{i=1}^n\frac{2i^3-3i}{n^4}=\frac{n^2(n+1)^2}{2n^4}-\frac{3n(n+1)}{2n^4}=\boxed{\dfrac{(n+1)(n^2+n-3)}{2n^3}}$