# How do you evaluate the sum represented by sum_(k=50)^300k^3?

Aug 29, 2015

The question is unclear, but if you meant ${\sum}_{n = 1}^{300} {k}^{3}$ with $k = \text{(constant) } 50$,
then
${\sum}_{n = 1}^{300} {k}^{3} = 300 \cdot {\left(50\right)}^{3} = 37500000$

#### Explanation:

If $k = 50$ then ${k}^{3} = 125000$ (a constant)

${\sum}_{n = 1}^{p} c$ for any constant $c$
$\textcolor{w h i t e}{\text{XXXX}} = p \cdot c$

Aug 29, 2015

#### Explanation:

This is the question. i don't understand this.

You can read this "the sum, from k equals 50 to 300, of k to the third"

or "the sum, of k to the third", from k equals 50 to 300"

There are other ways to say it as well, but these two are enough to start.

(for example you could say "from k = 50 to k= 300"

Aug 29, 2015

Use formula for ${\sum}_{k = 1}^{n} {k}^{3}$ to find ${\sum}_{k = 50}^{300} {k}^{3} = 2037021875$

#### Explanation:

Use ${\sum}_{k = 1}^{n} {k}^{3} = {\left({n}^{2} + n\right)}^{2} / 4$ (proved in another question recently).

Then:

${\sum}_{k = 50}^{300} {k}^{3} = {\sum}_{k = 1}^{300} {k}^{3} - {\sum}_{k = 1}^{49} {k}^{3}$

$= {\left({300}^{2} + 300\right)}^{2} / 4 - {\left({49}^{2} + 49\right)}^{2} / 4$

$= {90300}^{2} / 4 - {2450}^{2} / 4$

$= {45150}^{2} - {1225}^{2}$

$= 2038522500 - 1500625$

$= 2037021875$