Sigma? Sum!

Algebra Level 4

∑ n = 1 48 n 4 4 n 2 − 1 \large \sum_{n=1}^{48} \dfrac{n^4}{4n^2-1}

Round the summation above to the nearest integer. Submit your answer as the sum of digits of the integer you've found.


The answer is 23.

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Pankaj Joshi by Pankaj Joshi , 23, India

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2 solutions

Chew-Seong Cheong
May 21, 2015

∑ n = 1 48 n 4 4 n 2 − 1 = ∑ n = 1 48 16 n 4 − 1 + 1 16 ( 4 n 2 − 1 ) = 1 16 ∑ n = 1 48 ( 16 n 4 − 1 4 n 2 − 1 + 1 ( 2 n − 1 ) ( 2 n + 1 ) ) = 1 16 ∑ n = 1 48 ( 4 n 2 + 1 + 1 2 [ 1 2 n − 1 − 1 2 n + 1 ] ) = 1 4 ∑ n = 1 48 n 2 + 1 16 ∑ n = 1 48 1 + 1 32 ( ∑ n = 1 48 1 2 n − 1 − ∑ n = 2 49 1 2 n − 1 ) = 1 4 ( 48 × 49 × 97 6 ) + 1 16 ( 48 ) + 1 32 ( 1 1 − 1 97 ) = 9506 + 3 + 3 97 \begin{aligned} \sum_{n=1}^{48} {\frac{n^4}{4n^2-1}} & = \sum_{n=1}^{48} {\frac{16n^4-1+1}{16(4n^2-1)}} \\ & = \frac{1}{16} \sum_{n=1}^{48} {\left( \frac{16n^4-1}{4n^2-1} + \frac{1}{(2n-1)(2n+1)} \right)} \\ & = \frac{1}{16} \sum_{n=1}^{48} {\left( 4n^2+1 + \frac{1}{2}\left[ \frac{1}{2n-1}- \frac{1}{2n+1} \right] \right)} \\ & = \frac{1}{4} \sum_{n=1}^{48} { n^2} +\frac{1}{16} \sum_{n=1}^{48} {1} + \frac{1}{32}\left(\sum_{n=1}^{48} {\frac{1}{2n-1}} - \sum_{n=2}^{49} {\frac{1}{2n-1}}\right) \\ & = \frac{1}{4} \left( \frac{48\times 49\times 97}{6}\right) +\frac{1}{16} \left(48\right) + \frac{1}{32}\left(\frac{1}{1} - \frac{1}{97}\right) \\ & = 9506 + 3 + \frac{3}{97} \end{aligned}

Therefore, the nearest integer of the sum is 9509 9509 and the required answer is 9 + 5 + 0 + 9 = 23 9+5+0+9 = \boxed{23}

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Kindly review your solution (from 4th line), I can find various mistakes in it.

Abhishek Sharma - 5 years, 8 months ago

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Thanks. There were error starting from line 2.

Chew-Seong Cheong - 5 years, 8 months ago

LOL! When I saw this question, I first saw the first and the key step in my mind. But I did not reach there, nevertheless I actually followed another method reaching to the last step!

Kishore S. Shenoy - 5 years, 7 months ago

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