Notes: `sum_(i = 1)^6 (6i - 8i^3)` Find the sum using formulas for the sums of powers of integers.

Given: $\displaystyle{\sum_{{{i}={1}}}^{{6}}}{\left({6}{i}-{8}{{i}}^{{3}}\right)}$

$\displaystyle={\sum_{{{i}={1}}}^{{6}}}{6}{i}-{\sum_{{{i}={1}}}^{{6}}}{8}{{i}}^{{3}}$

$\displaystyle={6}{\sum_{{{i}={1}}}^{{6}}}{i}-{8}{\sum_{{{i}={1}}}^{{6}}}{{i}}^{{3}}$

Use the Sums of Powers of Integers to find the sums.

$\displaystyle{1}+{2}+{3}+{4}+\ldots+{n}=\frac{{{n}{\left({n}+{1}\right)}}}{{2}}$

$\displaystyle{{1}}^{{3}}+{{2}}^{{3}}+{{3}}^{{3}}+{{4}}^{{3}}+\ldots+{{n}}^{{3}}=\frac{{{{n}}^{{2}}{{\left({n}+{1}\right)}}^{{2}}}}{{4}}$

$\displaystyle={6}\frac{{{6}{\left({6}+{1}\right)}}}{{2}}-{8}\frac{{{{6}}^{{2}}{{\left({6}+{1}\right)}}^{{2}}}}{{4}}$

$\displaystyle={6}{\left({21}\right)}-{8}{\left({441}\right)}$

$\displaystyle=-{3402}$

The sum is -3402.

Notes