Verify the coefficient of $x^n$ from expansion is $\binom{2n+1}{n}$

Question

So... after 20 years I decided to mess with math again... And I now I feel I am either completely off, or missing a more explicit explanation on this one, can anyone give me a hand?

Verify that the coefficient of $x^n$ is $\binom {2n+1} {n}$ from the expansion

$(1+x)^{2n} + x(1+x)^{2n-1} + x^2(1+x)^{2n-2} + ... + x^n(1+x)^{n}$

[EDITED: removed proposed solution kept in the bottom for future reference]

I have realized the misspelling, that actually came into my calculations as well... SO i removed my erroneous stuff...

Thank you!

[reference: error... me being an idiot] From this I have used Newton's binom to reach the following:

$\sum^{2n}_{k=0}{\binom{2n}{k}}x^k(1+x)^{2n-k}$ which I transformed to, for $k=n$:

$\sum^{2n}_{k=n}{\binom{2n}{n}}x^n.(1+x)^{n} \iff$

$\sum^{2n}_{k=n}{\binom{2n}{n}}x^n.\sum^n_{j=0}\binom{n}{n}x^{n-j} \iff$

To me this means that each parcel is multiplied by the sequence $\brace {1,x^, x^2, ..., x^n} $

hence the combinations for the coefficient being $\binom{2n+1} {n}$

Thank you!

Answer 1

Critique of the Attempt

I don't understand how you got to $$ \sum_{k=0}^{2n}\binom{2n}{k}x^k(1+x)^{2n-k} $$ because that is $(1+2x)^{2n}$ and the coefficient of $x^n$ in that is $2^n\binom{2n}{n}$.

Unless I am mistaken, the second transformation is just plain wrong. You've gone to each term and replaced the index variable, $k$, with the constant, $n$. Then you removed half the terms by starting the summation at $n$ instead of $0$.

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Answer to the Question Asked

The question asked is $$ \begin{align} \left[x^n\right]\sum_{k=0}^nx^k(1+x)^{2n-k} &=\left[x^n\right](1+x)^n\sum_{k=0}^nx^k(1+x)^{n-k}\tag1\\ &=\left[x^n\right](1+x)^n\frac{(1+x)^{n+1}-x^{n+1}}{(1+x)-x}\tag2\\[6pt] &=\left[x^n\right]\left((1+x)^{2n+1}-(1+x)^nx^{n+1}\right)\tag3\\[9pt] &=\color{#090}{\left[x^n\right](1+x)^{2n+1}}-\color{#C00}{\left[x^n\right](1+x)^nx^{n+1}}\tag4\\[3pt] &=\color{#090}{\binom{2n+1}{n}}\tag5 \end{align} $$ Explanation:
$(1)$: pulled a factor of $(1+x)^n$ out front
$(2)$: formula for the sum of a geometric series
$(3)$: note that the denominator of the fraction is $1$, then distribute $(1+x)^n$
$(4)$: the "coefficient of" operator is real linear
$(5)$: apply the binomial theorem to the left term and note that the right term is a multiple of $x^{n+1}$

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Another Approach $$ \begin{align} \left[x^n\right]\sum_{k=0}^nx^k(1+x)^{2n-k} &=\left[x^n\right]\sum_{k=0}^{2n}x^k(1+x)^{2n-k}\tag6\\ &=\left[x^n\right]\frac{(1+x)^{2n+1}-x^{2n+1}}{(1+x)-x}\tag7\\[9pt] &=\color{#090}{\left[x^n\right](1+x)^{2n+1}}-\color{#C00}{\left[x^n\right]x^{2n+1}}\tag8\\[9pt] &=\color{#090}{\binom{2n+1}{n}}\tag9 \end{align} $$ Explanation:
$(6)$: include terms that are multiples of $x^{n+1}$
$(7)$: formula for the sum of a geometric series
$(8)$: denominator is $1$ and the "coefficient of" operator is real linear
$(9)$: apply the binomial theorem to the left term and the right term is $0$

Answer 2

Here is another take.

For the moment, let $y=1+x$. Then

$$ \begin{align} (1+x)^{2n} + x(1+x)^{2n-1} + \cdots + x^n(1+x)^{n} & = y^{2n} + xy^{2n-1} + \cdots + x^n y^{n} \\& = y^{n}(y^{n} + xy^{n-1} + \cdots + x^n ) \\& = y^{n}\frac{y^{n+1}-x^{n+1}}{y-x} \\& = y^{n}(y^{n+1}-x^{n+1}) \\& = y^{2n+1}-x^{n+1}y^{n} \\& = (1+x)^{2n+1}-x^{n+1}(1+x)^{n} \end{align} $$ Clearly, $x^n$ appears only in expansion of the first term and so its coefficient is $\binom{2n+1}{n}$.