Yes! You have seen it before

Algebra Level 4

$\large \displaystyle\sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}$

If the summation above equals to $\dfrac AB$ , where $A$ and $B$ are coprime positive integers, find $A+B$ .

The answer is 7.

2 solutions

Rishabh Jain
Feb 3, 2016

$\displaystyle \sum_{n=1}^{\infty}\frac{2}{(n+1)(n+2)}-\frac{1}{n(n+1)(n+2)}$ $\displaystyle \sum_{n=1}^{\infty}\frac{2}{(n+1)(n+2)}-\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$ $=2(\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n+1}-\dfrac{1}{n+2})-\dfrac{1}{2}(\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)})$ $~~~~~~~~~~~ ~~~~~~~~~~~ ~~~~~~ (\color{#302B94}{\small{ \text{Both are Telescopic series}}})$ $=\Large 1- \dfrac{1}{4} =\dfrac{3}{4}$ $\Large \therefore \color{forestgreen}{3+4=\color{#624F41}{\boxed{\color{#D61F06}{7}}}}$

Can you please state how you broke them into partial fractions

Chaitnya Shrivastava - 5 years, 4 months ago

Yes, of course..
Write first summation as: $2(\displaystyle \sum_{n=1}^{\infty}\dfrac{(n+2)-(n+1)}{(n+1)(n+2)})\\ =2(\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n+1}-\dfrac{1}{n+2})$ while for second summation write it as: $\frac{1}{2}(\displaystyle \sum_{n=1}^{\infty} \frac{(n+2)-(n)}{n(n+1)(n+2)})\\ \dfrac{1}{2}(\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)})$ Both summations are Telescopic series.. Feel free to ask further or for error rectification in the solution.. :- ]

Rishabh Jain - 5 years, 4 months ago

Thank you but did you think about it or is there any pattern or rule to it.

Chaitnya Shrivastava - 5 years, 4 months ago

@Chaitnya Shrivastava – Yup... Telescopic series.... you may click on View Wiki aside Discuss solution..

Rishabh Jain - 5 years, 4 months ago

@Rishabh Jain – No i was asking about breaking into partial fractions

Chaitnya Shrivastava - 5 years, 4 months ago

@Chaitnya Shrivastava – Maybe this could help..

Rishabh Jain - 5 years, 4 months ago

@Rishabh Jain – Thank you it helped a lot

Chaitnya Shrivastava - 5 years, 4 months ago

I feel this is a higher difficulty question. Read the wiki related to this buddy. You get more idea.

ASHWIN K - 5 years, 4 months ago

Thank you but which wiki should i refer to.

Chaitnya Shrivastava - 5 years, 4 months ago

Zk Lin
Feb 7, 2016

Express $\frac{2n-1}{n(n+1)(n+2)}$ as partial fractions:

$-\frac{1}{2n}+\frac{3}{n+1}-\frac{5}{2(n+2)}$

$=-\frac{1}{2n}+\frac{6}{2(n+1)}-\frac{5}{2(n+2)}$

$=-\frac{1}{2n}+\frac{1}{2(n+1)}+\frac{5}{2(n+1)}-\frac{5}{2(n+2)}$

$=-\frac{1}{2}(\frac{1}{n}-\frac{1}{n+1})+\frac{5}{2}(\frac{1}{n+1}-\frac{1}{n+2})$

Summing the telescopic series from $1$ to $\infty$ yields $-\frac{1}{2}+\frac{5}{4}=\frac{3}{4}$ . Therefore, the answer is $3+4= \boxed{7}$ .

Moderator note:

Simple standard approach.

For partial fractions with linear terms of multiplicity one in the denominator, you should CYA by saying "Using the cover up rule, ...."

How could you find a way to express the equation into partial fractions? Could you proof it? Thx

Victor Zhang - 5 years, 4 months ago

We write $\frac{2n-1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}$ . What we want to do is to find $A,B$ and $C$ which make the identity above holds true for all values of $n$ .

By cross multiplying and clearing denominators, we have:

$2n-1=A(n+1)(n+2)+B(n)(n+2)+C(n)(n+1)$

Since the equation above is an identity, we can substitute any values of $n$ we like. For convenience sake, we usually choose $n$ such that two of the three terms above equal $0$ . In this case, $n=0,-1,-2$ are excellent candidates.

By substituting $n=0$ , we get

$-1=2A$ , which implies that

$\boxed{A=-\frac{1}{2}}$

By substituting $n=-1$ , we get

$-3=-B$ , which implies that

$\boxed{B=3}$

By substituting $n=-2$ , we get

$-5=2C$ , which implies that

$\boxed{C=-\frac{5}{2}}$ .

And so we get the rational equation successfully written in partial fractions. Hope this clears your confusion!

ZK LIn - 5 years, 4 months ago

Yes! You have seen it before

The answer is 7.

2 solutions

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