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Algebra Level 4

If $\displaystyle \sum_{r=1}^\infty \dfrac r{r^4+r^2+1} = 1 - \dfrac1a$ , find the number of digits in $a^{2016}$ .

You are given the following approximations: $\log 10 = 1, \log 5 = 0.6990, \log 3 = 0.4771$ .

The answer is 607.

1 solution

Aniket Sanghi
Feb 1, 2016

First solve the series by factorising $r^4 + r^2 + 1 = (r^2 + r + 1)(r^2 - r + 1)$ then multiply numerator and denominator by 2 and write $2r = (r^2 + r+ 1)-(r^2 -r +1)$ and cancelling the similar terms from numerator and denominator.... now you could put values of r and identify the series... after getting a=2...to find no. of digits in ${ a }^{ 2016 } find \log { { a }^{ 2016 }}= 606.816$ .....hence number of digit = charateristic + 1........here.....606+1=607

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The answer is 607.

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