Why $ \sum _{k=0} ^{n-1} \binom{n-1}{k} x^{k+1} = \sum _{k=1} ^{n} \binom{n-1}{k-1} x^k $? I'm struggling to understand how to get from this:
$$ \sum _{k=0} ^{n-1} \binom{n-1}{k} x^{k+1}  $$
to this:
$$ \sum _{k=1} ^{n} \binom{n-1}{k-1} x^k  $$
I always have a problem understand the shifting of limits in general. Does there exist a guide/book somewhere?
 A: You should use a different variable name, here:
$$k':=k+1 \iff k=k'-1$$
In this way:
$$\sum _{k=0} ^{k=n-1} \binom{n-1}{k} x^{k+1}\tag{1}$$
(where I have cautiously written $k=...$ for the initial and the final values.)
becomes
$$\sum _{k'-1=0} ^{k'-1=n-1} \binom{n-1}{k'-1} x^{k'-1+1}$$
Otherwise said:
$$\sum _{k'=1} ^{k'=n} \binom{n-1}{k'-1} x^{k'}$$
Now, you are free to rename $k'$ as $k$...
A: The following representation might also be helpful. We can write
\begin{align*}
\sum_{k=0}^{n-1}\binom{n-1}{k}&=\sum_{\color{blue}{0\leq k\leq n-1}}\binom{n-1}{k}\tag{1}\\
&=\sum_{\color{blue}{1\leq k\leq n}}\binom{n-1}{k-1}=\sum_{k=1}^{n}\binom{n-1}{k-1}\tag{2}
\end{align*}
Some aspects:

*

*In (1) we use another common notation for the index region $0\leq k\leq n-1$. Here $k$ is the index variable going from $0$ to $n-1$.


*In (2) we shift the index variable by adding $1$ so that the index region becomes $1\leq k\leq n$. But we don't want to change the sum. We so have to compensate this by subtracting $1$ at each occurrence of $k$.


*This technique is independent from binomial coefficients. We have
\begin{align*}
  \sum_{k=0}^{n-1}a_k&=\sum_{\color{blue}{0\leq k\leq n-1}}a_k\\
  &=a_0+a_1+\cdots+a_{n-1}\\
  &=\sum_{\color{blue}{1\leq k\leq n}}a_{k-1}=\sum_{k=1}^na_{k-1}
\end{align*}
Hint: A thorough introduction in working with sums is given in chapter 2: Sums in Concrete Mathematics by R. L. Graham, D. Knuth and O. Patashnik.
