Binoharmonic summation in sequence and series

Question

Please consider this bino-harmonic summation:- nC0/a + nC0/a+d + nC0/ a+2d. . . . . . .nCn/a+nd I tried a lot but could not get anything any way all of it was failure..Now I am beginning to think that I am a real dumbass ,please help me solve this..

Answer 1

$$(1-x^b)^n=\sum_{r=0}^n\binom nr(-1)^n x^{bn}$$

Integrate both sides between $(0,1)$

Now if $\displaystyle I_n=\int_0^1(1-x^b)^ndx,I_0=1$

like series sum of binomial co-efficients, $$I_n=\dfrac{bn}{bn+1}I_{n-1}=\prod_{r=1}^n\dfrac{br}{br+1}$$

Put $b=\dfrac da$

Answer 2

I believe you set out to compute

$$ \frac1a\binom{n}{0} - \frac{1}{a+d}\binom{n}{1} + \cdots + (-1)^n\frac{1}{a + nd}\binom{n}{n}. $$

Consider the binomial expansion of $x^{a-1}(1-x^d)^n$:

$$ x^{a-1}(1-x^d)^n = x^{a-1}\left( \binom{n}{0} - \binom{n}{1}x^d + \cdots + (-1)^n\binom{n}{n}x^{nd} \right) \\ = \binom{n}{0}x^{a-1} - \binom{n}{1}x^{a+d-1} + \cdots + (-1)^n\binom{n}{n}x^{a+nd-1} $$

Integration w.r.t $x$,

$$ \int x^{a-1}(1-x^d)^n dx = \binom{n}{0}\frac{x^a}{a} - \binom{n}{1}\frac{x^{a+d}}{a+d} + \cdots + (-1)^n\binom{n}{n}\frac{x^{a+nd}}{a+nd} + c $$

Then, the series that you wanted to compute is given by

$$ \frac1a\binom{n}{0} - \frac{1}{a+d}\binom{n}{1} + \cdots + (-1)^n\frac{1}{a + nd}\binom{n}{n} = \int_0^1 x^{a-1}(1-x^d)^n dx. $$

Let $x^d = u$. Then $\frac{\text{du}}{\text{dx}} = dx^{d-1} \implies \text{dx} = \frac{\text{du}}{dx^{d-1}}$. The integral gets transformed to

$$ \frac1d\int_0^1 x^{a-d}(1-x^d)^n du = \frac1d\int_0^1 u^{a/d-1}(1-u)^n du $$

This integral is computed using the Gamma function as follows (See answer to this post):

$$ \frac1d\int_0^1 u^{a/d-1}(1-u)^n du = \frac1d\frac{\Gamma(a/d) \Gamma(n + 1)}{\Gamma(a/d + n + 1)}. $$