Why can the "i" represent two different things simultaneously?

Question

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Hi everyone, this question might be easy, but I'm still getting stuck. Can anyone explain to me why the 'i' can represent two different things simultaneously? Thank you very much!

The question is from the book: "A First Course in Probability" by Sheldon Ross.

The binomial theorem: $$(x+y)^n=\sum_{k=0}^n {n\choose k}x^ky^{n-k}$$

Proof of the Binomial Theorem by Induction: When n = 1, the Equation above reduces to $$x+y= {1\choose 0}x^0y^1+{1\choose 1}x^1 y^0=y+x$$

Assume Equation for n − 1. Now,

$$(x+y)^n=(x+y)(x+y)^{n-1}=(x+y)\sum_{k=0}^{n-1}x^k y^{n-1-k}=\sum_{k=0}^{n-1}{n-1 \choose k}x^{k+1}y^{n-1-k}+\sum_{k=0}^{n-1}{n-1 \choose k}x^k y^{n-k}$$

My question is here: Letting i=k+1 in the first sum and i=k in the second sum, we find that

$$(x+y)^n=\sum_{i=1}^{n}{n-1 \choose i-1}x^i y^{n-i}+\sum_{i=0}^{n-1}{n-1 \choose i}x^i y^{n-i}$$

Answer 1

The index variable $i$ is a so-called bound variable. This means that the scope (i.e. the range of validity) of the variable is determined by the sigma-operator $\sum$ and the operator precedence rules.

The following representations are valid \begin{align*} (x+y)^n&=\sum_{i=1}^{n}{n-1 \choose i-1}x^i y^{n-i}+\sum_{i=0}^{n-1}{n-1 \choose i}x^i y^{n-i}\\ &=\left(\color{green}{\sum_{i=1}^{n}{n-1 \choose i-1}x^i y^{n-i}}\right) +\left(\color{blue}{\sum_{i=0}^{n-1}{n-1 \choose i}x^i y^{n-i}}\right)\tag{1}\\ &=\sum_{i=1}^{n}{n-1 \choose i-1}x^i y^{n-i}+\sum_{k=0}^{n-1}{n-1 \choose k}x^k y^{n-k}\tag{2}\\ \end{align*}

Comment:

• In (1) we present the scope of each of the index variables somewhat more clearly by using parenthesis and the colors green and blue.

• In (2) we denote the index variable of the right-most sum with $k$.

Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.

Note: We have \begin{align*} \sum_{k=0}^{n-1}a_k&=a_0+a_1+\cdots+a_{n-1}\tag{3}\\ &=\sum_{k=1}^{n}a_{k-1}=\sum_{j=2}^{n+1}a_{j-2} \end{align*} The index shift at the sigma symbol $\sum$ is compensated at each occurrence of the index variable within the scope of the symbol. The index name $k, j, \ldots$ is not relevant, since it denotes always the right-hand sum at (3).