A crazy series, maybe

Calculus Level 4

( 3 3 ) + ( 5 3 ) 1 2 + ( 7 3 ) 1 2 2 + ( 9 3 ) 1 2 3 + ( 11 3 ) 1 2 4 + … = ? \binom{3}{3} + \binom{5}{3}\frac {1}{ 2 } + {7 \choose 3}\frac {1}{ 2^2 } + {9 \choose 3}\frac {1}{ 2^3 } + {11 \choose 3 }\frac {1}{ 2^4 } + \ldots = \ ?


The answer is 68.

Shaurya Gupta by shaurya gupta , 22, India

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1 solution

Shaurya Gupta
Nov 14, 2015

( 3 3 ) , ( 5 3 ) , ( 7 3 ) , . . . {3\choose 3}, {5 \choose 3}, {7 \choose 3}, ... are the alternative terms in the expansion of ( 1 + x ) − 4 (1+x)^{-4} .
( 1 − x ) − 4 + ( 1 + x ) − 4 2 = ( 3 3 ) + ( 5 3 ) x 2 + ( 7 3 ) x 4 + . . . \frac{(1-x)^{-4}+(1+x)^{-4}}{2} ={3\choose 3}+{5\choose 3}x^2+{7\choose 3}x^4+... Substitute x = 1 2 x=\sqrt{\frac{1}{2}} to get the required sum.

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Open it and observe that it is simply 1 3 ! ∑ n = 0 ∞ ( 2 n + 3 ) ( 2 n + 2 ) ( 2 n + 1 ) ( 1 2 ) 2 n \displaystyle \dfrac{1}{3!} \sum_{n=0}^{\infty}{(2n+3)(2n+2)(2n+1){\left(\frac{1}{\sqrt{2}}\right)}^{2n}}

= 1 6 d 3 d x 3 ( x 3 1 − x 2 ) ∣ x = 1 2 \displaystyle = \frac{1}{6} \dfrac{d^3}{dx^3}\left(\frac{x^3}{1-x^2}\right)|_{x=\frac{1}{\sqrt{2}}}

Kartik Sharma - 5 years, 7 months ago

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