Like here, how and why did we change it? When can we change this, when can’t we change the index cause I know it matters? I have seen many examples now and have no source of information regarding this
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Index like 0 to 1 – Ashish Raj Shukla Dec 23 '17 at 08:05
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Try rewriting both sums, you will see that they are exactly the same. They are just using a 'different' dummy variable. – Karn Watcharasupat Dec 23 '17 at 08:13
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If I have understood it correctly, the confusion you are having is how to write $$\sum_{r = 0}^{n}r \,^{n}C_{r} = n \sum_{r = 1}^{n}\,^{n-1}C_{r-1}.$$
The proof is here:
\begin{align} \sum_{r = 0}^{n}r \, ^{n}C_{r} & = \underbrace{0 \cdot \,^{n}C_{0}}_{=\, 0} + 1 \cdot \,^{n}C_{1} + \cdots + n \cdot \,^{n}C_{n} \\ & = 1 \cdot \,^{n}C_{1} + \cdots + n \cdot \,^{n}C_{n} = \sum_{r = 1}^{n}r\,^{n}C_{r} \\ & = \sum_{r = 1}^{n} \left[r \times \dfrac{n!}{(n-r)! \, r!} \right] = \sum_{r = 1}^{n} \left[r \times \dfrac{n \times (n-1)!}{(n-r)! \times \, r \times (r-1)!} \right]\\ & = n \left[ \sum_{r = 1}^{n}\dfrac{(n-1)!}{((n-1) - (r-1))!\,(r-1)!}\right] \\ & = n \sum_{r = 1}^{n} \, ^{n-1}C_{r-1} \end{align}
Nash J.
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