Summation

**Punch** · June 9th 2011, 07:36 PM

Find $\sum_{n=1}^n(\frac{1}{4r^2-8r+3})$ given that
$\frac{2}{4r^2-8r+3}=\frac{1}{2r-3}-\frac{1}{2r-1}$

$\sum_{n=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{n= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})$

$=\frac{1}{2}\sum_{n=1}^n(\frac{1}{2r-3})-\frac{1}{2}\sum_{n= 1}^n(\frac{1}{2r-1})$

$=\frac{1}{2}[\frac{1}{\frac{n}{2}(-2+(n-1)(2))}]-\frac{1}{2}[\frac{1}{\frac{n}{2}(2-(n-1)(2))}]$

$=\frac{1}{2}[\frac{1}{\frac{n}{2}(2n-4)}-\frac{1}{\frac{n}{2}(4-2n)}]$

$= \frac{1}{2}[\frac{1}{\frac{n}{2}(2n-4)}+\frac{1}{\frac{n}{2}(2n-4)}]$

$=\frac{1}{\frac{n}{2}(2n-4)}$

$=\frac{1}{n(n-2)}$

but ans is $-\frac{n}{2n-1}$

**Sudharaka** · June 9th 2011, 08:26 PM

Originally Posted by Punch

Find $\sum_{n=1}^n(\frac{1}{4r^2-8r+3})$ given that
$\frac{2}{4r^2-8r+3}=\frac{1}{2r-3}-\frac{1}{2r-1}$

$\sum_{n=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{n= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})$

$=\frac{1}{2}\sum_{n=1}^n(\frac{1}{2r-3})-\frac{1}{2}\sum_{n= 1}^n(\frac{1}{2r-1})$

$=\frac{1}{2}[\frac{1}{\frac{n}{2}(-2+(n-1)(2))}]-\frac{1}{2}[\frac{1}{\frac{n}{2}(2-(n-1)(2))}]$

$=\frac{1}{2}[\frac{1}{\frac{n}{2}(2n-4)}-\frac{1}{\frac{n}{2}(4-2n)}]$

$= \frac{1}{2}[\frac{1}{\frac{n}{2}(2n-4)}+\frac{1}{\frac{n}{2}(2n-4)}]$

$=\frac{1}{\frac{n}{2}(2n-4)}$

$=\frac{1}{n(n-2)}$

but ans is $-\frac{n}{2n-1}$

You have taken, $\sum_{r=1}^{n}\frac{1}{2r-3}=\frac{1}{\frac{n}{2}(-2+(n-1)(2))}$ .

This is incorrect (substitute n=2 and the right hand side becomes undefined).

Much easier is to treat the summation without breaking it apart into two pieces.

You have to evaluate, $\sum_{r=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{r= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})$

When r=1; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{-1}-\frac{1}{1}$

When r=2; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{1}-\frac{1}{3}$

When r=3; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{3}-\frac{1}{5}$
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When r=n-1; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{2n-5}-\frac{1}{2n-3}$

When r=n; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{2n-3}-\frac{1}{2n-1}$

Adding all these together you will get the correct answer.

**Punch** · June 9th 2011, 08:52 PM

Originally Posted by Sudharaka

You have taken, $\sum_{r=1}^{n}\frac{1}{2r-3}=\frac{1}{\frac{n}{2}(-2+(n-1)(2))}$ .

This is incorrect (substitute n=2 and the right hand side becomes undefined).

Much easier is to treat the summation without breaking it apart into two pieces.

You have to evaluate, $\sum_{r=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{r= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})$

When r=1; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{-1}-\frac{1}{1}$

When r=2; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{1}-\frac{1}{3}$

When r=3; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{3}-\frac{1}{5}$
.
.
.
.
.
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When r=n-1; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{2n-5}-\frac{1}{2n-3}$

When r=n; $\frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{2n-3}-\frac{1}{2n-1}$

Adding all these together you will get the correct answer.

$\sum_{r=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{r= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})$

$=-2+\frac{2}{3}+\frac{2}{15}+\frac{2}{35}$

but this is neither a GP nor AP

**Punch** · June 9th 2011, 08:55 PM

wait, i think i see it, is this the method of differences?

**TheCoffeeMachine** · June 9th 2011, 08:59 PM

Originally Posted by Punch

$\frac{1}{2}\sum_{n=1}^n(\frac{1}{2r-3})-\frac{1}{2}\sum_{n= 1}^n(\frac{1}{2r-1})$ [/TEX]

At this stage, you can just map the summands to each other, by, say, replacing r with r+1 for the first one:

$\displaystyle\begin{aligned} S & = \frac{1}{2}\sum_{1 \le r \le n}\frac{1}{2r-3}-\frac{1}{2}\sum_{1 \le r \le n }^n\frac{1}{2r-1} \\& = \frac{1}{2}\sum_{1 \le r+1 \le n}\frac{1}{2(r+1)-3}-\frac{1}{2}\sum_{1 \le r \le n }^n\frac{1}{2r-1} \\& = \frac{1}{2}\sum_{0 \le r \le n-1}\frac{1}{2r-1}-\frac{1}{2}\sum_{1 \le r \le n }^n\frac{1}{2r-1} \\& = \frac{1}{2}\left(\frac{1}{2(0)-1}\right) -\frac{1}{2}\left(\frac{1}{2(n)-1}\right) + \frac{1}{2}\sum_{1 \le r \le n}\frac{1}{2r-1}-\frac{1}{2}\sum_{1 \le r \le n }^n\frac{1}{2r-1} \\& = \frac{1}{2}\bigg(\frac{1}{1-2n}-1\bigg).\end{aligned}$

Doing the above might seem bit hard/weird at first, but after few times it becomes so natural that you will never again want to write out terms and stuff! :]

**Sudharaka** · June 9th 2011, 09:11 PM

Originally Posted by Punch

wait, i think i see it, is this the method of differences?

Yes. Every term cancels with a term succeeding it. i.e: -1/1 cancels with 1/1, -1/3 cancels with 1/3 etc. Only the terms, -1 and $-\frac{1}{2n-1}$ will be left.

Hence $\sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)$

**Soroban** · June 9th 2011, 10:37 PM

Hello, Punch!

$\text{Find }\,\sum_{n=1}^n\left(\frac{1}{4r^2-8r+3}\right)\,\text{ given that: }\:\frac{1}{4r^2-8r+3}\:=\:\frac{1}{2}\left(\frac{1}{2r-3}-\frac{1}{2r-1}\right)$

Write out the terms . . .

$S \;=\; \frac{1}{2}\sum^n_{r=1}\left(\frac{1}{2r-3} - \frac{1}{2r-1}\right)$

. . $=\;\frac{1}{2}\bigg[\left(-\frac{1}{1} - \frac{1}{1}\right) + \left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \hdots + \left(\frac{1}{2n-3} - \frac{1}{2n-1}\right)\bigg]$

Inside the brackets, all terms cancel out except the first and last terms.

. . $S \;=\;\frac{1}{2}\left(-1 - \frac{1}{2n-1}\right) \;=\;\frac{1}{2}\left(\frac{-2n}{2n-1}\right)$

Therefore: . $S \;=\;-\frac{n}{2n-1}$

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