I am attempting to put:
$$a\sum_{n=0}^{\infty} \sum_{k=0}^{n} [{n \choose k}b^{k}c^{n-k} (2+5k+3(n-k))]$$
where a=1/6, b=1/3, and c=1/2 into a closed form, but I haven't been able to figure out a way to do it. Is it possible to be done, and if so, how do I go about doing it?
Thanks!
On
We have that
$$\sum_{k=0}^{n} [{n \choose k}b^{k}c^{n-k} (2+5k+3(n-k))]=$$
$$=2\sum_{k=0}^{n} {n \choose k}b^{k}c^{n-k} +5\sum_{k=0}^{n} k{n \choose k}b^{k}c^{n-k}+3\sum_{k=0}^{n} (n-k){n \choose k}b^{k}c^{n-k}$$
and
$$2\sum_{k=0}^{n} {n \choose k}b^{k}c^{n-k}=2(b+c)^n$$
$$5\sum_{k=0}^{n} k{n \choose k}b^{k}c^{n-k}=5b\sum_{k=0}^{n} k{n \choose k}b^{k-1}c^{n-k}=5nb(b+c)^{n-1}$$
$$3\sum_{k=0}^{n} (n-k){n \choose k}b^{k}c^{n-k}=3\sum_{k=0}^{n} {n \choose n-k}b^{n-k}c^{k}=3(b+c)^n$$
therefore
$$a\sum_{n=0}^{\infty} \sum_{k=0}^{n} [{n \choose k}b^{k}c^{n-k} (2+5k+3(n-k))]=5a\sum_{n=0}^{\infty} (b+c)^n+nb(b+c)^{n-1}$$
$$ a \sum_{n=0}^{\infty} \sum_{k=0}^{n} [{n \choose k}b^{k}c^{n-k} (2))] = a\sum_{n=0}^{\infty} 2 (b+c)^n $$ $$ a \sum_{n=0}^{\infty} \sum_{k=0}^{n} [{n \choose k}b^{k}c^{n-k} (2k)] = a\sum_{n=0}^{\infty} 2 b (b+c)^{n-1} n $$ $$ a \sum_{n=0}^{\infty} \sum_{k=0}^{n} [{n \choose k}b^{k}c^{n-k} (3n))] = a\sum_{n=0}^{\infty} 3 (b+c)^n n $$
The first and third are the binomial formula. The second comes from "the derivative trick".
We know how to $\sum_k \binom{n}{k}b^k c^{n-k}$, but how do we $\sum_k \binom{n}{k} k b^k c^{n-k}$? Notice that \begin{align*} \frac{\mathrm{d}}{\mathrm{d}b} \binom{n}{k}b^k c^{n-k} &= \binom{n}{k} k b^{k-1} c^{n-k} \text{, so} \\ \sum_{k=0}^n \binom{n}{k} k b^{k} c^{n-k} &= b\sum_{k=0}^n \binom{n}{k} k b^{k-1} c^{n-k} \\ &= b \frac{\mathrm{d}}{\mathrm{d}b} \sum_{k=0}^n\binom{n}{k}b^k c^{n-k} \\ &= b \frac{\mathrm{d}}{\mathrm{d}b}(b+c)^n \\ &= b (b+c)^{n-1} n \text{.} \end{align*}
We used the binomial formula to discharge the sum. Note that your original form suggests using $\frac{\mathrm{d}}{\mathrm{d}c}$ to get the "$3(n-k)$" rather than just distributing as I did. Also, additional derivatives can be used to deal with higher powers of the index and integration to deal with negative powers of the index. (I've sometimes wondered if fractional integration or differentiation would make, for instance, $\sum_k \binom{n}{k} \sqrt{k} b^k c^{n-k}$ either easy or a terrible mess.)
Of the three resulting sums, the first is a geometric series in $n$ and the last two are geometric series after applying the derivative trick described previously, so you can evaluate them, add them up, and find that your sum is $\frac{a(3b+c+2)}{(b+c-1)^2}$.