Solve summation expression

Question

For a probability problem, I ended up with the following expression $$\sum_{k=0}^nk\ \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k$$ Using Mathematica I've found that the result should be $\frac{n}{3}$. However, I have no idea how to get there. Any ideas?

Answer 1

\begin{align*} \sum_{k=0}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k &=\sum_{k=0}^n k\frac{n!}{k!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\ &=\sum_{k=0}^n \frac{n!}{(k-1)!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\ &=\frac{n}{3}\sum_{k=0}^n \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}\left(\frac{2}{3}\right)^{(n-1)-(k-1)}\left(\frac{1}{3}\right)^{k-1}\\ &=\frac{n}{3}\sum_{k=0}^n \binom{n-1}{k-1}\left(\frac{2}{3}\right)^{(n-1)-(k-1)}\left(\frac{1}{3}\right)^{k-1}\\ &=\frac{n}{3}\left(\frac{2}{3}+\frac{1}{3}\right)^{n-1}\\ &=\frac{n}{3}. \end{align*} The second to last line is a result of the binomial theorem.

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Edit: It was pointed out that I need to be careful when $k=0$. I also made a mistake in applying the binomial theorem. Here is a revised proof. Note that when $k=0$, the term of the sum is $0$, so it is the same as starting at $k=1$.

\begin{align*} \sum_{k=0}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k &= \sum_{k=1}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\ &= \sum_{k=1}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\ &=\sum_{k=1}^n k\frac{n!}{k!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\ &=\sum_{k=1}^n \frac{n!}{(k-1)!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\ &=\frac{n}{3}\sum_{k=1}^n \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}\left(\frac{2}{3}\right)^{(n-1)-(k-1)}\left(\frac{1}{3}\right)^{k-1}\\ &=\frac{n}{3}\sum_{k=1}^n \binom{n-1}{k-1}\left(\frac{2}{3}\right)^{(n-1)-(k-1)}\left(\frac{1}{3}\right)^{k-1}\\ &=\frac{n}{3}\sum_{k=0}^{n-1} \binom{n-1}{k}\left(\frac{2}{3}\right)^{(n-1)-k}\left(\frac{1}{3}\right)^{k}\\ &=\frac{n}{3}\left(\frac{2}{3}+\frac{1}{3}\right)^{n-1}\\ &=\frac{n}{3}. \end{align*}

Answer 2

It is the expected value of $B(n,\frac{1}{3})$, where $B(n,p)$ is a binomial distribution with parameters $n\in \mathbb{N}$ and $p\in [0,1]$. Let $X\sim B(n,\frac{1}{3})$, then $$ \sum_{k=0}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^{k} = E[X]= \frac{n}{3}. $$

Answer 3

Consider the function $$f\left(x\right)=\left(\frac{2}{3}+x\right)^{n}. $$ By the binomial theorem we have $$f\left(x\right)=\sum_{k=0}^{n}\dbinom{n}{k}\left(\frac{2}{3}\right)^{n-k}x^{k} $$ so if we take the derivative and we multiply by $x$ we have $$xf'\left(x\right)=\sum_{k=0}^{n}k\dbinom{n}{k}\left(\frac{2}{3}\right)^{n-k}x^{k} $$ so $$\sum_{k=0}^{n}k\dbinom{n}{k}\left(\frac{2}{3}\right)^{n-k}x^{k}=nx\left(\frac{2}{3}+x\right)^{n-1} $$ now take $x=\frac{1}{3}$.

Answer 4

Hint: $$k\binom{n}{k}=n\binom{n-1}{k-1}$$

Answer 5

You can solve the following question in two different ways:

• Using binomial distribution
• Using the direct approach

When rolling a fair $6$-sided die $n$ times, what is the expected number of times we get "1" or "2"?

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Using binomial distribution, we split it into disjoint events and then add up their probabilities:

• The probability of getting "1" or "2" in a single roll is $\frac13$
• The probability of getting "1" or "2" in exactly $k$ out of $n$ rolls is $\binom{n}{k}\cdot\left(\frac13\right)^{k}\cdot\left(1-\frac13\right)^{n-k}$
• The expected number of times we get "1" or "2" in $n$ rolls is $\sum\limits_{k=0}^{n}k\cdot\binom{n}{k}\cdot\left(\frac13\right)^{k}\cdot\left(1-\frac13\right)^{n-k}$

Using the direct approach, we expect to get "1" or "2" exactly $\frac13$ of the time, i.e., in $\frac{n}{3}$ out of $n$ rolls.

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Therefore:

$$\sum\limits_{k=0}^{n}k\cdot\binom{n}{k}\cdot\left(\frac13\right)^{k}\cdot\left(1-\frac13\right)^{n-k}=\frac{n}{3}$$

Answer 6

Just some binomial coefficient massage and the binomial formula: $$ \sum_{k=0}^nk\binom nkx^{n-k}y^k =\sum_{k=1}^nn\binom{n-1}{k-1}x^{n-k}y^k =ny(x+y)^{n-1} $$ Now plug in $x=\frac23$ and $y=\frac13$.