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Algebra Level 4

$\large{ \dfrac{1}{1^4 + 1^2 + 1} + \dfrac{2}{2^4 + 2^2 + 1} + \ldots + \dfrac{2016}{2016^4 + 2016^2 + 1} ~=~?}$

Give your answer to 3 decimal places.

The answer is 0.500.

1 solution

Akshat Sharda
Jan 4, 2016

$\begin{aligned} S & = \displaystyle \sum^{2016}_{n=1}\frac{n}{n^4+n^2+1} \\ & = \displaystyle \sum^{2016}_{n=1} \frac{n}{(n^2+1+n)(n^2+1-n)} \\ & = \frac{1}{2} \displaystyle \sum^{2016}_{n=1} \left(\frac{1}{n^2+1-n} - \frac{1}{n^2+1+n}\right) \\ & = \frac{1}{2} \left(1-\frac{1}{3}+\frac{1}{3} - \frac{1}{7} + \frac{1}{7} -\frac{1}{13} + \ldots +\frac{1}{4062241}-\frac{1}{4066273} \right) \\ & = \frac{1}{2} \left(1-\frac{1}{4066273} \right) \\ & = \frac{4066272}{8132546} \\ \Rightarrow S & = \boxed{0.499} \end{aligned}$

I did same. (+1)

Dev Sharma - 5 years, 5 months ago

Well i did not bothered to calculate the 2016th term! .( i knew it would be very small ) hence entered 0.5 and got correct

Prakhar Bindal - 5 years, 5 months ago

To three places it must be 0.500

Rishabh Jain - 5 years, 5 months ago

I got 0.499

Akshat Sharda - 5 years, 5 months ago

0.49999988.... As the no. of terms in the summation grows their contribution to the sum decreases significantly till 2016, that contribution has tended to negligible.

Rishabh Jain - 5 years, 5 months ago

@Rishabh Jain – It goes like this -

0.49999987703727713313887188587682...

Dev Sharma - 5 years, 5 months ago

If the sum goes to infinity then the value is 0.500

Akshat Sharda - 5 years, 5 months ago

Let's telescope

The answer is 0.500.

1 solution

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