Notes: `sum_(n = 1)^20 (n^3 - n)` Find the sum using formulas for the sums of powers of integers.

Thursday, June 5, 2014

$\displaystyle{\sum_{{{n}={1}}}^{{20}}}{\left({{n}}^{{3}}-{n}\right)}$ Find the sum using formulas for the sums of powers of integers.

You need to evaluate the given sum using formulas for the sums of powers of integers, such that:

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{\left({{n}}^{{3}}-{n}\right)}={\sigma_{{{n}={1}}}^{{20}}}{{n}}^{{3}}-{\sigma_{{{n}={1}}}^{{20}}}{n}$

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{{n}}^{{3}}={{1}}^{{3}}+{{2}}^{{3}}+\ldots.+{{20}}^{{3}}$

You need to remember that $\displaystyle{\sigma_{{{n}={1}}}^{{n}}}{{k}}^{{3}}={{\left(\frac{{{n}{\left({n}+{1}\right)}}}{{2}}\right)}}^{{2}}$

Replacing 20 for n yields:

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{{n}}^{{3}}={{\left(\frac{{{20}{\left({20}+{1}\right)}}}{{2}}\right)}}^{{2}}$

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{{n}}^{{3}}={100}\cdot{{21}}^{{2}}$

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{n}={1}+{2}+\ldots+{20}$

You need to remember that $\displaystyle{\sigma_{{{n}={1}}}^{{n}}}{k}={\left(\frac{{{n}{\left({n}+{1}\right)}}}{{2}}\right)}$

Replacing 20 for n yields:

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{n}=\frac{{{20}{\left({20}+{1}\right)}}}{{2}}$

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{n}={10}\cdot{21}$

Evaluating the sum, yields:

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{\left({{n}}^{{3}}-{n}\right)}={100}\cdot{{21}}^{{2}}-{10}\cdot{21}$

Factoring out $\displaystyle{10}\cdot{21}$ yields:

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{\left({{n}}^{{3}}-{n}\right)}={10}\cdot{21}\cdot{\left({10}\cdot{21}-{1}\right)}$

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{\left({{n}}^{{3}}-{n}\right)}={210}\cdot{209}$

$\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{\left({{n}}^{{3}}-{n}\right)}={43890}$

Hence, evaluating the given sum using formulas for the sums of powers of integers, yields $\displaystyle{\sigma_{{{n}={1}}}^{{20}}}{\left({{n}}^{{3}}-{n}\right)}={210}\cdot{209}={43890}.$

Notes

Thursday, June 5, 2014

$\displaystyle{\sum_{{{n}={1}}}^{{20}}}{\left({{n}}^{{3}}-{n}\right)}$ Find the sum using formulas for the sums of powers of integers.

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