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Calculus Level 4

( ∑ n = 0 ∞ 1 5 n ( 2 n n ) ) 2 = ? \large \left(\sum_{n=0}^\infty \dfrac1{5^n}\dbinom{2n}n \right)^2 = \, ?

Notation : ( M N ) \binom MN denotes the binomial coefficient , ( M N ) = M ! N ! ( M − N ) ! \binom MN = \frac{M!}{N!(M-N)!} .


The answer is 5.

Hamza A by Hamza A , 19, USA

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1 solution

Chew-Seong Cheong
Jun 21, 2016

The generating function of central binomial coefficient is:

∑ n = 0 ∞ ( 2 n n ) x n = 1 1 − 4 x Putting x = 1 5 ⟹ ∑ n = 0 ∞ ( 2 n n ) 1 5 n = 5 ( ∑ n = 0 ∞ ( 2 n n ) 1 5 n ) 2 = 5 \begin{aligned} \sum_{n=0}^\infty {2n \choose n}x^n & = \frac 1{\sqrt{1-4x}} \quad \quad \small \color{#3D99F6}{\text{Putting }x = \frac 15} \\ \implies \sum_{n=0}^\infty {2n \choose n} \frac 1{5^n} & = \sqrt 5 \\ \left( \sum_{n=0}^\infty {2n \choose n} \frac 1{5^n} \right)^2 & = \boxed{5} \end{aligned}

3 Helpful 1 Interesting 0 Brilliant 0 Confused

How did you wrote the first line?

Aakash Khandelwal - 4 years, 11 months ago

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You can find the reference here .

Chew-Seong Cheong - 4 years, 11 months ago

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