Use the properties of summation to evaluate the sum

**calculus123** · December 2nd 2012, 10:53 PM

Could someone tell me how to solve this?

**MarkFL** · December 3rd 2012, 12:33 AM

I would use the formulas:

$\sum_{k=1}^n(1)=n$

$\sum_{k=1}^n(k)=\frac{n(n+1)}{2}$

$\sum_{k=1}^n(k^2)=\frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^n(k^3)=\frac{n^2(n+1)^2}{4}$

and use the property $\sum_{k=1}^n(a\cdot f(k)\pm b\cdot g(k))=a\cdot\sum_{k=1}^n(f(k))\pm b\cdot\sum_{k=1}^n(g(k))$ to rewrite the sum first, then use the above formulas.

**calculus123** · December 4th 2012, 09:52 AM

Is this correct?

**Plato** · December 4th 2012, 10:40 AM

Originally Posted by calculus123

Is this correct?

NO! It is not.

It should be $\sum\limits_{k = 1}^{10} {k^3 } - 2\sum\limits_{k = 1}^{10} {k^2 } + 5\sum\limits_{k = 1}^{10} k - \sum\limits_{k = 1}^{10} {(1)}$

See reply #2 to finish.

**calculus123** · December 5th 2012, 02:57 PM

I have spent all day today trying to solve it...I can't. Can someone please solve it and show me to do it?

**MarkFL** · December 5th 2012, 03:17 PM

$\sum_{k=1}^{10}(k^3)-2\sum_{k=1}^{10}(k^2)+5\sum_{k = 1}^{10}(k)-\sum_{k = 1}^{10}(1)=$

$\frac{10^2\cdot11^2}{4}-2\cdot\frac{10\cdot11\cdot21}{6}+5\cdot\frac{10 \cdot11}{2}-10=$

$3025-770+275-10=2520$

**Plato** · December 5th 2012, 03:19 PM

Originally Posted by calculus123

I have spent all day today trying to solve it...I can't. Can someone please solve it and show me to do it?

You really need to have a live tutor.

Originally Posted by MarkFL2

I would use the formulas:
$\sum_{k=1}^n(1)=n$
$\sum_{k=1}^n(k)=\frac{n(n+1)}{2}$
$\sum_{k=1}^n(k^2)=\frac{n(n+1)(2n+1)}{6}$
$\sum_{k=1}^n(k^3)=\frac{n^2(n+1)^2}{4}$

Use the above:

$\sum\limits_{k = 1}^{10} {k^3 } - 2\sum\limits_{k = 1}^{10} {k^2 } + 5\sum\limits_{k = 1}^{10} k - \sum\limits_{k = 1}^{10} {(1)}$

Now plug in:
$\frac{10^2(10+1)^2}{4}-2\left(\frac{10(10+1)(2\cdot 10+1)}{6}\right)+5\left( \frac{10(10+1)}{2}\right)- 10$

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