To understand why

$\sum_{k=0}^{n-1} \binom{n}{k} x^{k+1} y^{n-k} = \sum_{k=1}^{n} \binom{n}{k-1} x^k y^{n-k+1},$

let’s delve into the reasoning step by step:

1. Left-hand side: $\sum_{k=0}^{n-1} \binom{n}{k} x^{k+1} y^{n-k}.$ Here, $x^{k+1}$ and $y^{n-k}$ are written in terms of $k$, with the summation starting at $k = 0$ and going to $n-1$. Notice that the $k+1$ in the exponent of $x$ and $n-k$ in the exponent of $y$ suggest a reindexing might be necessary.

2. Reindexing $k+1$: Let’s set $j = k + 1$. Then:

   • When $k = 0$, $j = 1$.
   • When $k = n-1$, $j = n$.

   Under this change of index, we rewrite $k = j - 1$, and the summation becomes: $\sum_{j=1}^{n} \binom{n}{j-1} x^j y^{n-(j-1)}.$

3. Simplify powers of $y$: Since $n - (j-1) = n - j + 1$, the expression simplifies further to: $\sum_{j=1}^{n} \binom{n}{j-1} x^j y^{n-j+1}.$

4. Rename the index: Renaming $j$ back to $k$, we get: $\sum_{k=1}^{n} \binom{n}{k-1} x^k y^{n-k+1}.$

This matches the right-hand side of the equation. The equality holds because of the reindexing step and the combinatorial property of binomial coefficients.

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Would you like further clarification or examples to solidify this understanding? Here are five follow-up questions to explore:

1. Can we interpret this result combinatorially using binomial coefficients?
2. How does this equality change if we swap $x$ and $y$?
3. How does reindexing affect similar summations in calculus or combinatorics?
4. Can you expand and simplify specific cases, e.g., $n = 3$?
5. What role does the binomial coefficient shifting $\binom{n}{k-1}$ play?

Tip: Always check bounds and indices carefully when reindexing summations.