series sum of binomial co-efficients

Question

$C_r$ stands for $_nC_r$

We have to show that

$ \frac{C_0}{1} -\frac{C_1}{5} + \frac{C_2}{9} +\ldots+ (-1)^n\cdot\frac{C_n}{4n+1} = \frac{4^n\cdot n!}{1.5.9\ldots(4n+1)}$

What I have done :

$\int_0^1(1-x^4)^ndx = \int_0^1(C_0 -C_1x^4+C_2x^8-\ldots +(-1)^nC_nx^{4n})dx$

$ I = $ Left hand side of what we have to prove . Now I have to just evaluate $I$ but I am stuck badly . Help me out in finding the integral .

Related : http://math.stackexchange.com/questions/437523/proving-binomial-idenity-without-calculus — lab bhattacharjee, Dec 04 '13 at 03:42

Norbert · Answer 1 · 2013-12-03T13:10:59.937

2

$$ \int_0^1(1-x^4)^ndx=\{t=1-x^4\}=\frac{1}{4}\int_0^1 t^n (1-t)^{-3/4}dt=B\left(n+1,\frac{1}{4}\right)=\frac{\Gamma(n+1)\Gamma(1/4)}{\Gamma(n+5/4)} $$

edited Dec 03 '13 at 13:10

answered Dec 03 '13 at 12:59

Norbert

58,398

@Norbet what is $B(n+1,1/4)$ ? – abkds Dec 03 '13 at 13:00
This is the beta function – Norbert Dec 03 '13 at 13:01

score 2 · Accepted Answer · answered Dec 03 '13 at 13:17

Set :

$$I_{n}=\int_0^1(1-x^4)^n$$

Integrating by part, we have:

$$I_{n}=\int_0^1(1-x^4)^ndx=x(1-x^4) \mid_{0}^1- \int_{0}^{1}x .n(1-x^4).(-4).x^3dx=4n\int_{0}^{1}x^4(1-x^4)^{n-1}dx=-4n\int_{0}^{1}(1-x^4)(1-x^4)^{n-1}dx+4n\int_{0}^{1}(1-x^4)^{n-1}dx=-4nI_n+4nI_{n-1}$$

Which gives: $$I_{n}=\dfrac{4.n}{(1+4n)}.I_{n-1}= \frac{4^n\cdot n!}{1.5.9\ldots(4n+1)}$$

series sum of binomial co-efficients

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