KVPY #9

Algebra Level 4

If a = ∑ r = 1 ∞ 1 r 2 \displaystyle a = \sum_{r=1}^\infty \frac 1{r^2} and b = ∑ r = 1 ∞ 1 ( 2 r − 1 ) 2 \displaystyle b = \sum_{r=1}^\infty \frac 1{(2r-1)^2} , find 3 a b \dfrac {3a}b .

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The answer is 4.

Rahil Sehgal by Rahil Sehgal , 20, India

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2 solutions

Rahil Sehgal
Mar 27, 2017

13 Helpful 0 Interesting 0 Brilliant 0 Confused

This is the Basel problem

Vijay Simha - 4 years, 2 months ago
James Pohadi
Apr 1, 2017

b = ∑ r = 1 ∞ 1 ( 2 r − 1 ) 2 = 1 1 2 + 1 3 2 + 1 5 2 + . . . \displaystyle b = \sum_{r=1}^\infty \dfrac 1{(2r-1)^2} = \dfrac{1}{1^{2}} +\dfrac{1}{3^{2}}+ \dfrac{1}{5^{2}} + ...

a = ∑ r = 1 ∞ 1 r 2 = 1 1 2 + 1 2 2 + 1 3 2 + . . . a = ( 1 1 2 + 1 3 2 + 1 5 2 + . . . ) + 1 4 ( 1 1 2 + 1 3 2 + 1 5 2 + . . . ) + 1 16 ( 1 1 2 + 1 3 2 + 1 5 2 + . . . ) + . . . a = b + 1 4 b + 1 16 b + . . . a = b 1 − 1 4 = b 3 4 3 a b = 4 \displaystyle a = \sum_{r=1}^\infty \dfrac 1{r^2} = \dfrac{1}{1^{2}} +\dfrac{1}{2^{2}}+ \dfrac{1}{3^{2}} + ... \\ \displaystyle a = (\dfrac{1}{1^{2}} +\dfrac{1}{3^{2}}+ \dfrac{1}{5^{2}} + ...) + \dfrac{1}{4} (\dfrac{1}{1^{2}} +\dfrac{1}{3^{2}}+ \dfrac{1}{5^{2}} + ...) + \dfrac{1}{16} (\dfrac{1}{1^{2}} +\dfrac{1}{3^{2}}+ \dfrac{1}{5^{2}} + ...)+... \\ a=b+\dfrac{1}{4}b+\dfrac{1}{16}b+... \\ a=\dfrac{b}{1-\dfrac{1}{4}}=\dfrac{b}{\dfrac{3}{4}} \\ \dfrac{3a}{b}=4

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