Notes: `sum_(j = 1)^10 (3 - 1/2 j + 1/2 j^2)` Find the sum using formulas for the sums of powers of integers.

Monday, June 20, 2016

$\displaystyle{\sum_{{{j}={1}}}^{{10}}}{\left({3}-\frac{{1}}{{2}}{j}+\frac{{1}}{{2}}{{j}}^{{2}}\right)}$ 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:

$\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{\left({3}-{\left(\frac{{1}}{{2}}\right)}{j}+{\left(\frac{{1}}{{2}}\right)}{{j}}^{{2}}\right)}={\sigma_{{{j}={1}}}^{{103}}}-{\left(\frac{{1}}{{2}}\right)}{\sigma_{{{j}={1}}}^{{10}}}{j}+{\left(\frac{{1}}{{2}}\right)}{\sigma_{{{j}={1}}}^{{10}}}{{j}}^{{2}}$

$\displaystyle{\sigma_{{{j}={1}}}^{{103}}}={3}+{3}+..+{3}={10}\cdot{3}={30}$

$\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{j}={1}+{2}+..+{10}$

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

Replacing 10 for j yields:

$\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{j}={10}\frac{{{10}+{1}}}{{2}}={5}\cdot{11}$

$\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{{j}}^{{2}}={{1}}^{{2}}+{{2}}^{{2}}+\ldots+{{10}}^{{2}}$

You need to remember that $\displaystyle{\sigma_{{{n}={1}}}^{{n}}}{{k}}^{{2}}={\left(\frac{{{n}{\left({n}+{1}\right)}{\left({2}{n}+{1}\right)}}}{{6}}\right)}$

Replacing 10 for n yields:

$\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{{j}}^{{2}}={\left(\frac{{{10}{\left({10}+{1}\right)}{\left({2}\cdot{10}+{1}\right)}}}{{6}}\right)}$

$\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{{j}}^{{2}}=\frac{{{10}\cdot{11}\cdot{21}}}{{6}}={5}\cdot{11}\cdot{7}={35}\cdot{11}$

Evaluating the sum, yields:

$\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{\left({3}-{\left(\frac{{1}}{{2}}\right)}{j}+{\left(\frac{{1}}{{2}}\right)}{{j}}^{{2}}\right)}={30}-{\left(\frac{{1}}{{2}}\right)}\cdot{5}\cdot{11}+{\left(\frac{{1}}{{2}}\right)}\cdot{35}\cdot{11}$

$\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{\left({3}-{\left(\frac{{1}}{{2}}\right)}{j}+{\left(\frac{{1}}{{2}}\right)}{{j}}^{{2}}\right)}=\frac{{{60}-{55}+{385}}}{{2}}={195}$

Hence, evaluating the given sum using formulas for the sums of powers of integers, yields $\displaystyle{\sigma_{{{j}={1}}}^{{10}}}{\left({3}-{\left(\frac{{1}}{{2}}\right)}{j}+{\left(\frac{{1}}{{2}}\right)}{{j}}^{{2}}\right)}={195}.$

Notes

Monday, June 20, 2016

$\displaystyle{\sum_{{{j}={1}}}^{{10}}}{\left({3}-\frac{{1}}{{2}}{j}+\frac{{1}}{{2}}{{j}}^{{2}}\right)}$ Find the sum using formulas for the sums of powers of integers.

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