The advisor asks to verify that the coefficient of $$x^n$$ in the development of: $$(1+x)^{2n}+x(1+x)^{2n−1}+x2(1+x)^{2n−2}+......+x^n(1+x)^n$$ is equal to $$\binom{2n+1}{n}$$
I tried for summations but not. I did with the denominator change too, but I can not even,how can i match the 2 questions
On
Your sum is $$S =\sum_{k=0}^n x^k(1+x)^{2n-k} $$
Developping, you get : $$S = \sum_{k=0}^n x^k \sum_{j=0}^{2n-k} {2n-k \choose j} x^j = \sum_{k=0}^n \sum_{j=0}^{2n-k} {2n-k \choose j} x^{j+k} = \sum_{N=0}^{2n} \sum_{k=0}^{n} {2n-k \choose N-k} x^{N} $$
You see that the coefficient in $n$ is equal to $$\sum_{k=0}^n {2n-k \choose n-k} = \sum_{j=0}^n {n+j \choose j} = {2n+1 \choose n}$$
On
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain \begin{align*} \color{blue}{[x^n]}&\color{blue}{\left((1+x)^{2n}+x(1+x)^{2n-1}+\cdots+x^n(1+x)^n\right)}\\ &=[x^n]\sum_{j=0}^nx^j(1+x)^{2n-j}\\ &=\sum_{j=0}^n[x^{n-j}](1+x)^{2n-j}\tag{1}\\ &=\sum_{j=0}^n[x^j](1+x)^{n+j}\tag{2}\\ &=[x^0](1+x)^n\sum_{j=0}^n\left(\frac{1+x}{x}\right)^j\tag{3}\\ &=[x^0](1+x)^n\frac{\left(\frac{1+x}{x}\right)^{n+1}-1}{\frac{1+x}{x}-1}\tag{4}\\ &=[x^0](1+x)^n\frac{(1+x)^{n+1}-x^{n+1}}{x^n}\\ &=[x^n](1+x)^{2n+1}-[x^{-1}](1+x)^n\tag{5}\\ &\,\,\color{blue}{=\binom{2n+1}{n}} \end{align*}
and the claim follows.
Comment:
In (1) we use the formula $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
In (2) we change the order of summation $j\to n-j$.
In (3) we factor out terms independent of $j$ and use also the formula from comment (1).
In (4) we apply the finite geometric series formula.
In (5) we use again the formula from comment (1).
In (6) we select the coefficient of $x^n$, the other term does not contribute anything.
On
You can explicitly evaluate your sum and extract the coefficient afterward. Recall that $$(y-x) (y^n + xy^{n-1} + \cdots + x^{n-1}y + x^n) = y^{n+1} - x^{n+1}$$ Letting $y = 1+x$ (so $y-x = 1$) gives
$$(1+x)^n + x(1+x)^{n-1} + \cdots + x^{n-1}(1+x) + x^n = (1+x)^{n+1} - x^{n+1}.$$ Then multiplying by $(1+x)^n$ gives your expression
$$(1+x)^{2n} + x(1+x)^{2n-1} + \cdots + x^n (1+x)^n = (1+x)^{2n+1} - (1+x)^n x^{n+1}.$$
The coefficient of $x^n$ of $(1+x)^nx^{n+1}$ is $0$, while the coefficient of $x^n$ in $(1+x)^{2n+1}$ is $${ 2n+1 \choose n}$$ by the binomial theorem.
$$(1+x)^{2n}+x(1+x)^{2n-1}+x^2(1+x)^{2n-2}+...+x^n(1+x)^n$$ $$=\sum_{k=0}^{2n}\binom{2n}{k}x^k+x\sum_{k=0}^{2n-1}\binom{2n-1}{k}x^k+...+x^n\sum_{k=0}^{n}\binom{n}{k}x^k$$ for the $x^n$ coefficient in this summation we need to add each of the $x^n$ coefficients in individual terms. This gives us $$\binom{2n}{n}+\binom{2n-1}{n-1}+...+\binom{2n-n}{n-n}=\sum_{k=0}^{n}\binom{2n-k}{n-k}=\binom{2n+1}{n}$$