coefficient $x ^ n$ in development 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
 A: 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.
A: $$(1+x)^{2n}+x(1+x)^{2n-1}+x^2(1+x)^{2n-2}+...+x^n(1+x)^n$$
$$=\sum_{k=0}^{2n}\binom{2n}{k}x^k+x\sum_{k=0}^{2n-1}\binom{2n-1}{k}x^k+...+x^n\sum_{k=0}^{n}\binom{n}{k}x^k$$
for the $x^n$ coefficient in this summation we need to add each of the $x^n$ coefficients in individual terms. This gives us
$$\binom{2n}{n}+\binom{2n-1}{n-1}+...+\binom{2n-n}{n-n}=\sum_{k=0}^{n}\binom{2n-k}{n-k}=\binom{2n+1}{n}$$
A: 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
$$\sum_{k=0}^n  {2n-k \choose n-k} = \sum_{j=0}^n {n+j \choose j} = {2n+1 \choose n}$$
A: 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.
