Use the properties of summation to evaluate the sum

• Dec 2nd 2012, 10:53 PM
calculus123
Use the properties of summation to evaluate the sum
http://i50.tinypic.com/2v9u4qf.jpg
Could someone tell me how to solve this?
• Dec 3rd 2012, 12:33 AM
MarkFL
Re: Use the properties of summation to evaluate the sum
I would use the formulas:

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

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

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

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

and use the property $\displaystyle \sum_{k=1}^n(a\cdot f(k)\pm b\cdot g(k))=a\cdot\sum_{k=1}^n(f(k))\pm b\cdot\sum_{k=1}^n(g(k))$ to rewrite the sum first, then use the above formulas.
• Dec 4th 2012, 09:52 AM
calculus123
Re: Use the properties of summation to evaluate the sum
• Dec 4th 2012, 10:40 AM
Plato
Re: Use the properties of summation to evaluate the sum
Quote:

Originally Posted by calculus123

NO! It is not.

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

• Dec 5th 2012, 02:57 PM
calculus123
Re: Use the properties of summation to evaluate the sum
I have spent all day today trying to solve it...I can't. Can someone please solve it and show me to do it?
• Dec 5th 2012, 03:17 PM
MarkFL
Re: Use the properties of summation to evaluate the sum
$\displaystyle \sum_{k=1}^{10}(k^3)-2\sum_{k=1}^{10}(k^2)+5\sum_{k = 1}^{10}(k)-\sum_{k = 1}^{10}(1)=$

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

$\displaystyle 3025-770+275-10=2520$
• Dec 5th 2012, 03:19 PM
Plato
Re: Use the properties of summation to evaluate the sum
Quote:

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.

Quote:

Originally Posted by MarkFL2
I would use the formulas:
$\displaystyle \sum_{k=1}^n(1)=n$
$\displaystyle \sum_{k=1}^n(k)=\frac{n(n+1)}{2}$
$\displaystyle \sum_{k=1}^n(k^2)=\frac{n(n+1)(2n+1)}{6}$
$\displaystyle \sum_{k=1}^n(k^3)=\frac{n^2(n+1)^2}{4}$

Use the above:

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

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