Proving binomial sum (checking my answer)

Question

$\sum_{k=1}^{20} \binom{18}{k+3}(k+2)(k+1)3^{-3-k}$

$3^{-3}\sum_{k=1}^{16} \binom{16}{k+1}3^{-k}$

Substitute $t=k+1$

$3^{-3}\sum_{t=2}^{16} \binom{16}{t}3^{1-t}$

$3^{-2}\sum_{t=0}^{16} \binom{16}{t}(\frac{1}{3})^{t}-1-\frac{16}{3}$

$ 3^{-2}[(\frac{1}{3}+1)^{16}-1-\frac{16}{3}] = 3^{-2}[\frac{4}{3}^{16}-1-\frac{16}{3}]$

Is this correct?

Answer 1

It is not correct: the first step is invalid. Here’s one way to get started. First shift the index:

$$\begin{align*} \sum_{k=1}^{20}\binom{18}{k+3}(k+2)(k+1)3^{-3-k}&=\sum_{k=4}^{18}\binom{18}k(k-1)(k-2)3^{-k}\\ &=\sum_{k=4}^{18}\binom{18}k\big(k(k-1)-2k+2\big)3^{-k}\,. \end{align*}$$

Then

$$\binom{18}kk(k-1)=18\cdot17\cdot\binom{16}{k-2}$$

and

$$\binom{18}kk=18\binom{17}{k-1}$$

so you can split it into three nicer sums; one of them, for instance, is

$$18\cdot17\sum_{k=4}^{18}\binom{16}{k-2}3^{-k}\,.$$