Binoharmonic summation in sequence and series 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..

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