Different ways of computing $\sum_{n=0}^{\infty}\binom{n+7}{n}\left(\frac{1}{3}\right)^{n}$ Calculate $\sum_{n=0}^{\infty}\binom{n+7}{n}\left(\frac{1}{3}\right)^{n}$ in various ways.$$$$
one more question: how about $\sum_{n=0}^{\infty}n\binom{n+7}{n}\left(\frac{1}{3}\right)^{n}$ ?
 A: Hint: there's a more general theorem $$\frac{1}{(1-z)^{k+1}}=\sum_{n=0}^\infty \binom{n+k}{n}z^n$$ with $|z|<1$
To get the second question, take the derivative of this and multiply by $z$, yielding:
$$\frac{(k+1)z}{(1-z)^{k+2}} = \sum_{n=0}^\infty n\binom{n+k}{n} z^n$$
You can prove the first identity combinatorially by asking, "how many ways can I write $n$ as the sum of an ordered sequence of $k+1$ non-negative integers?" The "stars and bars" theorem says that number is $\binom{n+k}{k}=\binom{n+k}{n}$. But if we think of the left side as the product of $k+1$ copies of $\frac{1}{1-z}$ we see that the resulting coefficient of $z^n$ in that product is just this count, so both sides are equal.
Or you can prove that first identity by induction on $k$ starting at $k=0$, and then multiplying both sides by $\frac{1}{1-z}=1+z+z^2\dots$ to go from $k$ to $k+1$, using the relationship:
$$\sum_{m=0}^n \binom{m+k}{m} = \binom{n+k+1}{n}$$
Or you could prove it by taking the $k$th derivative of $\frac{1}{1-z}=\sum_{n=0}^\infty z^n$, as others have done.
Or you could prove it by using the general binomial theorem.
For the second problem, an alternative approach is to note that:
$$n\binom{n+k}{n}=(k+1)\binom{n+k}{n-1}$$
A: $$\binom{n+7}{n}x^n=\frac{1}{7!}\frac{d^7}{dx^7}{x^{n+7}}$$
So you have $$\left[\frac{d^7}{dx^7}\sum_{n=0}^\infty\frac{1}{7!}x^{n+7}\right]_{x\to1/3}$$
which is $$\left[\frac{d^7}{dx^7}\frac{1}{7!}x^7\sum_{n=0}^\infty x^{n}\right]_{x\to1/3}$$
which is  $$\left[\frac{d^7}{dx^7}\left(\frac{1}{7!}x^7\frac1{1-x}\right)\right]_{x\to1/3}$$
which is, according to the product rule for higher order derivatives,  $$\left[\sum_{k=0}^7\binom{7}{k}\frac{d^k}{dx^k}\left(\frac{1}{7!}x^7\right)\frac{d^{7-k}}{dx^{7-k}}\left(\frac1{1-x}\right)\right]_{x\to1/3}$$
which is  $$\left[\sum_{k=0}^7\binom{7}{k}\frac{1}{(7-k)!}x^{7-k}(7-k)!(1-x)^{-1-(7-k)}\right]_{x\to1/3}$$
which is  $$\sum_{k=0}^7\binom{7}{k}(1/3)^{7-k}(2/3)^{k-8}$$
or just  $$\frac{3}{2^8}\sum_{k=0}^7\binom{7}{k}2^k$$
where we see something familiar from the binomial theorem in the sum, and have at long last  $$\frac{3}{2^8}(1+2)^7$$ aka $$\left(\frac{3}{2}\right)^8$$

For your second question,
$$\begin{align}n\binom{n+7}{n}x^n&=\frac{1}{7!}\frac{d^7}{dx^7}n{x^{n+7}}\\
&=\frac{1}{7!}\frac{d^7}{dx^7}\left(x^8\frac{d}{dx}x^n \right)\\
\end{align}$$
A: Remember the generalized binomial theorem:
$$
   \left( 1 + q x \right)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} q^n x^n
$$
Also, for negative integral values of the exponent $\alpha = -m$:
$$
   \binom{-m}{n} = \frac{\prod_{k=0}^{n-1} (-m-k)}{n!} = (-1)^n  \frac{\prod_{k=0}^{n-1} (m+k)}{n!} \stackrel{k \to n-1-k}{=} (-1)^n \frac{\prod_{k=0}^{n-1} (n+m-1-k)}{n!} = (-1)^n \binom{n+m-1}{n}
$$
Thus
$$
   \left(1+q x\right)^{-m} = \sum_{n=0}^\infty \binom{n+m-1}{n} (-1)^n q^n x^n
$$ 
Now set $\alpha = -m=-8$ and $q=-\frac{1}{3}$.
A: Here is a different way (the steps would need some justification as we are manipulation infinite power series, but they are all valid for $|x| \lt 1$)
Let
$$f_{r}(x) = \sum_{n=0}^{\infty} \binom{n+r}{r} x^n $$
Then we have that 
$$f_r(x) - xf_r(x) = 1 + \sum_{n=1}^{\infty} \left(\binom{n+r}{r} - \binom{n+r-1}{r}\right)x^n$$
Using the identity
$$\binom{n}{k} - \binom{n-1}{k} = \binom{n-1}{k-1}$$
we get
$$f_r(x) - xf_r(x) = 1 + \sum_{n=1}^{\infty} \binom{n+r-1}{r-1}x^n = f_{r-1}(x)$$
Thus
$$f_r(x) = \dfrac{f_{r-1}(x)}{1-x}$$
Since we have that $f_0(x) = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ we have that
$$f_r(x) = \dfrac{1}{(1-x)^{r+1}}$$
A similar approach will work for your other question.
