# "Rewrite using summation notation"

• Jan 29th 2009, 10:45 AM
fattydq
"Rewrite using summation notation"
I understand how to do summation notation and the formula for it (n(n+1))/2 and how that differs if n has an exponent in it, but what if you're asked to rewrite a problem in summation notation?

For example "Rewrite 1+2+3...+100 in summation notation"
and just so I have a clear understanding let's also say
"Rewrite 1^2+2^2+3^2...+100^2 in summation notation"

• Jan 29th 2009, 11:37 AM
chabmgph
Quote:

Originally Posted by fattydq
I understand how to do summation notation and the formula for it (n(n+1))/2 and how that differs if n has an exponent in it, but what if you're asked to rewrite a problem in summation notation?

For example "Rewrite 1+2+3...+100 in summation notation"
and just so I have a clear understanding let's also say
"Rewrite 1^2+2^2+3^2...+100^2 in summation notation"

For the first one, let $i$ be your counter,
so it would be $\sum_{i=1}^{100}i$

For the second one, still let $i$ be your counter,
so it would be $\sum_{i=1}^{100}i^2$
• Jan 29th 2009, 11:42 AM
fattydq
Quote:

Originally Posted by chabmgph
For the first one, let $i$ be your counter,
so it would be $\sum_{i=1}^{100}i$

For the second one, still let $i$ be your counter,
so it would be $\sum_{i=1}^{100}i^2$

oh wow, that's really all they mean you think? And what does the i=1 thing mean, is that the interval between the values or something? So if it went 1+3+5+7 i would equal 2?
• Jan 29th 2009, 01:24 PM
Jester
Quote:

Originally Posted by fattydq
oh wow, that's really all they mean you think? And what does the i=1 thing mean, is that the interval between the values or something? So if it went 1+3+5+7 i would equal 2?

Your right, the common jump is two. If you write the terms like

$
\begin{array}{c}
(1+2\cdot 0) + (1+2\cdot 1) + (1+2\cdot 2) + (1 + 2\cdot 3) + \cdots
\end{array}$

then

$\sum_{i=0}^4 1 + 2i$.
• Jan 29th 2009, 01:33 PM
fattydq
Quote:

Originally Posted by danny arrigo
Your right, the common jump is two. If you write the terms like

$
\begin{array}{c}
(1+2\cdot 0) + (1+2\cdot 1) + (1+2\cdot 2) + (1 + 2\cdot 3) + \cdots
\end{array}$

then

$\sum_{i=0}^4 1 + 2i$.

gotcha, thanks a bunch man