Value of sum $\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^{n-k}$ .

Question

$\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^{n-k}$ is equal to

$1.$ $1$.

$2.$ $x$.

$3.$ $x^2$

$4.$ $x^n$.

According to me $B_n(x)=\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^k$ form polynomials called Bernstein polynomial tends to continuous function $x$ because $f(\frac{k}{n})=\frac{k}{n}$. So limit of $B_n(x)$ will tend to $x$, but in question it is asked without limit. So what will be correct answer and why ?. Thank you for solution.

Answer 1

I'm assuming you meant $\sum_{k=0}^{n}\dbinom{n}{k}\dfrac{k}{n}x^{k}(1-x)^{n-k} $
Note that $k\dbinom{n}{k} = n\dbinom{n-1}{k-1} \implies \dfrac{k}{n}\dbinom{n}{k} = \dbinom{n-1}{k-1}$ $$ B_n(x)=\sum_{k=0}^{n}\dbinom{n}{k}\frac{k}{n}x^{k}(1-x)^{n-k} $$ $$ = x \sum_{k=1}^{n} \dbinom{n-1}{k-1} x^{k-1}(1-x)^{n-k} $$ $$ = x(x+1-x)^{n-1}$$ $$ =x $$

Answer 2

You can see this :

$$ \sum_{k=0}^n { n \choose k}\dfrac{k}{n}x^k(1-x)^{n-k}= \mathbb{E}(X/n) $$ where $X$ follows a Bernoulli of parameter $x$.

It follows from here that :

$$ \sum_{k=0}^n { n \choose k}\dfrac{k}{n}x^k(1-x)^{n-k}=nx/n=x $$