# Simplify this equation.

Can I simplify or approximate this equation without sigma and combination?

\begin{align}
\sum_{i = 0}^n (-1)^i {n \choose i} \frac{{d+1}}{d(di + 1)}
\end{align}

#### Solutions Collecting From Web of "Simplify this equation."

By the binomial theorem, $$(1-x^d)^n = \sum_{i=0}^n (-1)^i \binom{n}{i} x^{di}.$$

Now note that the integral $\int_{0}^1 x^{di}\,dx = \frac{1}{di+1}$. So

$$\int_0^1 (1-x^d)^n\,dx = \sum_{i=0}^n (-1)^i \binom{n}{i} \frac{1}{di+1},$$
which means that your sum is equal to $\frac{d+1}{d} \int_0^1 (1-x^d)^n\,dx.$

Finally, as pointed out by Jack D’Aurizio, we can express this in terms of the gamma function by making the substitution $z=x^d$, $dz = dx^{d-1} dx$ which changes the integral to

$$\frac{d+1}{d^2} \int_0^1 (1-z)^n z^{1/d – 1} dz = \frac{d+1}{d^2} B(n+1,\frac1d) = \frac{d+1\, \Gamma(n+1)\, \Gamma(\frac1d)}{d^2 \,\Gamma(n+1+\frac1d)} = \frac{\Gamma(n+1)\, \Gamma(2 + \frac1d)}{ \,\Gamma(n+1+\frac1d)}.$$

where $B$ is the classical beta function.