A lusty summation

Algebra Level 4

$\large \displaystyle \sum_{n=1}^{\infty} \left(\dfrac{4}{n^2+4n+3}-9^{2-n}4^{n+1}\right)$

If the value of above summation can be expressed in the form $-\dfrac{m}{n}$ , where $m$ and $n$ are coprime positive integers, find $m-n$ .

The answer is 3848.

1 solution

Rishabh Jain
Jan 18, 2016

Summation can be broken into two brackets: $\large (4 \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{(n+1)(n+3)})-(324 \displaystyle \sum_{n=1}^{\infty}(\dfrac{4}{9})^n)$ $\small{\color{#EC7300}{\text{(First bracket is a telescopic series while}}}$ $\small{\color{#EC7300}{\text{second one is sum of a infinite GP)}}}$ $=2(\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n+1}-\dfrac{1}{n+3})-324(\displaystyle \sum_{n=1}^{\infty}\dfrac{\frac{4}{9}}{1-\frac{4}{9}})$ $=2(\frac{1}{2}+\frac{1}{3})-324(\frac{4}{5})$ $=-\dfrac{3863}{15}$ $\Large\color{magenta}{m-n=3848}$

Just one typo -

$\displaystyle \sum_{n=1}^{\infty} \left(\frac{4} {n^2+4n+3} \right)=4 \displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{n^2+4n+3}\right)=\frac{4}{2} \displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{(n+1)}-\frac{1}{(n+3)}\right)$

Akshat Sharda - 5 years, 4 months ago

Fixed............. and solution updated..

Rishabh Jain - 5 years, 4 months ago

Thank you for a great solution and error rectification!

Rohit Udaiwal - 5 years, 4 months ago

@Rohit Udaiwal – No problem.... :)

Rishabh Jain - 5 years, 4 months ago

Rishabh Cool's Cool Solution.

Kushagra Sahni - 5 years, 4 months ago

$\Large\color{forestgreen}{Thanks}$

Rishabh Jain - 5 years, 4 months ago

What is your surname? Which class you in? Which state do you belong to? I am operating brilliant from the app so can't see your profile.

Kushagra Sahni - 5 years, 4 months ago

@Kushagra Sahni – My name is Jain...Rishabh Jain .... 12....Ajmer(Rajasthan)

Rishabh Jain - 5 years, 4 months ago

Same approach.

Niranjan Khanderia - 2 years, 9 months ago

A lusty summation

The answer is 3848.

1 solution

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