I am working on an inclusion-exclusion problem. At some point, I've calculated the probability of a term as:
$$\sum_{k=1}^{l}\binom l k \cdot \left(\frac 3 n\right)^k\left(1-\frac 3 n\right)^{l-k}$$
But of course, binomial theorem requires the index of summation to start at $0$.
I tried to change index by letting $j = k -1 \Leftrightarrow k = j + 1$ and working with that, but I'm unable to simplify the $\binom n {j + 1}$ to $\binom n j$.
Here is what I got so far:
$$\frac {\frac 3 n}{1 - \frac 3 n}\sum_{j=0}^l\frac{l-(j+1)}{j+1}\binom n j\left(\frac 3 n\right)^j\left(1-\frac 3 n\right)^{l-j}$$
But of course, I can't take out the large fraction out of the summation because it relies on $j$. How can I apply binomial theorem here?
It is convenient to apply the binomial theorem as follows: \begin{align*} \color{blue}{\sum_{k=1}^{l}}&\color{blue}{\binom{l}{k}\left(\frac{3}{n}\right)^k\left(1-\frac{3}{n}\right)^{l-k}}\\ &=\sum_{k=\color{blue}{0}}^{l}\binom{l}{k}\left(\frac{3}{n}\right)^k\left(1-\frac{3}{n}\right)^{l-k}-\binom{l}{0}\left(\frac{3}{n}\right)^0\left(1-\frac{3}{n}\right)^{l-0}\\ &=\left(\frac{3}{n}+\left(1-\frac{3}{n}\right)\right)^l-\left(1-\frac{3}{n}\right)^l\\ &\,\,\color{blue}{=1-\left(1-\frac{3}{n}\right)^l} \end{align*}