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Evaluate the following sum:

$$\sum_{k=0}^n\binom{n}{k}\frac{k}{n}x^k(1-x)^{n-k}$$

This almost looks like the Binomial Theorem, except there's that pesky little $\frac{k}{n}$ term in there that changes things up a bit. I'm not quite sure what steps to take from here. How do you evaluate this sum?

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2 Answers 2

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Use $\frac kn\binom nk=\binom{n-1}{k-1}$ to get $$\sum_{k=1}^n\binom{n-1}{k-1}x^k(1-x)^{n-k}$$ which can be summed by the binomial theorem.

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$$\sum_{k=0}^n\binom{n}{k}\frac{k}{n}x^k(1-x)^{n-k}=\sum_{k=1}^n\frac{n!k}{k!(n-k)!n}x^k(1-x)^{n-k}=\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}x^k(1-x)^{n-k}=\sum_{k=1}^n\binom{n-1}{k-1}x^k(1-x)^{n-k}=\dfrac{x}{1-x}\sum_{k=0}^{n-1}\binom{n-1}{k}x^k(1-x)^{n-k}=\dfrac{x}{1-x}\cdot(1-x+x)^{n-1}=\dfrac{x}{1-x}$$

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  • $\begingroup$ it's a polynomial. $\endgroup$
    – Angina Seng
    Commented Feb 11, 2018 at 20:40
  • $\begingroup$ Yes and it directly yields to correct answer $\endgroup$
    – Mostafa Ayaz
    Commented Feb 11, 2018 at 20:50

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