Pi By What!

Calculus Level 4

r = 0 1 16 r 2 + 16 r + 3 \large \sum^{\infty}_{r=0} \dfrac{1}{16r^{2} + 16r + 3} The summation above evaluates to a π b \dfrac{a\pi}{b} for coprime positive integers a , b a,b . Calculate a + b \sqrt{ a+b} .


The answer is 3.000.

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1 solution

Let S = r = 0 T r S= \sum^{\infty}_{r=0}T_{r} T r = 1 16 r 2 + 16 r + 3 T_{r} = \dfrac{1}{16r^{2} + 16r + 3}

T r = 1 ( 4 r + 1 ) ( 4 r + 3 ) T_{r}= \dfrac{1}{\left(4r+1\right)\left(4r+3\right)}

T r = 1 2 ( 1 4 r + 1 1 4 r + 3 ) T_{r}= \dfrac {1}{2}\left( \dfrac{1}{4r+1} - \dfrac{1}{4r+3}\right)

S = 1 2 ( 1 1 3 + 1 5 1 7 . . . . . ) S= \dfrac{1}{2}\left(1- \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} .....\right) Consider f ( x ) = 1 1 + x 2 = ( 1 + x 2 ) 1 f\left(x\right) = \dfrac {1}{1+ x^{2}} = \left(1+x^{2}\right)^{-1}

Expanding f f by binomial theorem for any rational index OR you may also use Taylor's series expansion f ( x ) = 1 x 2 + x 4 x 6 + . . . f\left(x\right) = 1- x^{2} +x^{4} - x^{6} +... Integrating, tan 1 ( x ) = x x 3 3 + x 5 5 . . . . . \tan^{-1}\left(x\right) = x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - ..... Put x = 1 x=1 to obtain π 4 = 2 S \dfrac{\pi}{4} = 2S Therefore a = 1 , b = 8 a=1 , b=8 Which gives a + b = 3.00 \sqrt{ a+b} = 3.00

NOTE:

1) you may have used the series of tan 1 ( x ) \tan^{-1}\left(x\right) directly,which I didn't (for the sake of completeness of solution.

2)please reshare if you like the problem ,and follow for more problems

Nice problem and solution. I just recognized that k = 1 ( 1 ) k + 1 2 k 1 = π 4 \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{2k - 1} = \dfrac{\pi}{4}

and continued from there.

Brian Charlesworth - 6 years, 1 month ago

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Thanks sir,I am glad you liked it :)

Siddharth Bhatnagar - 6 years, 1 month ago

Nice use of Leibniz formula.

Swapnil Das - 5 years, 7 months ago

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