If x+y=1, then ∑r=0nrnCrxr·ym=

∑r=0nrCr nxr·yn-r=∑r=0nr·n!(n-r)!r!xr·yn-r=nx∑r=1nCr-1 n-1xr-1·yn-1-r-1=nx(y+x)n-1=nx

If x+y=1, then ∑r=0nrnCrxr·ym=

∑r=0nrCr nxr·yn-r=∑r=0nr·n!(n-r)!r!xr·yn-r=nx∑r=1nCr-1 n-1xr-1·yn-1-r-1=nx(y+x)n-1=nx

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$If x + y = 1, then \sum_{r = 0}^{n} r^{n} C_{r} x^{r} \cdot y^{m} =$

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answer is C.

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Detailed Solution

$\begin{array}{l} \sum_{r = 0}^{n} r {}^{n}C_{r} x^{r} \cdot y^{n - r} = \sum_{r = 0}^{n} r \cdot \frac{n!}{(n - r)! r!} x^{r} \cdot y^{n - r} \\ = n x \sum_{r = 1}^{n} {}^{n - 1}C_{r - 1} x^{r - 1} \cdot y^{(n - 1) - (r - 1)} = n x (y + x)^{n - 1} = n x \end{array}$