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Coefficient of $x^3$ in $(1+x^2)(1+x)^{100}$

Coefficient of $x^{10}$ in $(1+x)^{10}(1-x)^{10}$

Coefficient of $x^n$ in $\dfrac{2+x}{2-x}$.

Any help on these would be appreciated.

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It would help if you would tell us what you know and show us what you've tried. Do you know the binomial theorem? Do you see how to use it to expand $(1+x)^{100}$? Etc., etc. –  Gerry Myerson Mar 22 '12 at 5:03
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For the second one, note that we are looking at $(1-x^2)^{10}$ and use the known expansion of $(1+w)^{10}$. For the third one, rewrite as $\frac{1+x/2}{1-x/2}$ and use the known expansion of $\frac{1}{1-w}$. –  André Nicolas Mar 22 '12 at 5:42

2 Answers 2

  1. There are two ways to obtain $x^3$ in $(1+x^2)(1+x)^{100}$: when you multiply $1$ on the first binomial by the $x^3$-term in the expansion of $(1+x)^{100}$, and when you multiply $x^2$ by the $x$-term in the expansion of $(1+x)^{100}$.

  2. You obtain $x^{10}$ by multiplying the $x^i$ term in the expansion of $(1+x)^{10}$ with the $x^{10-i}$ term in the expansion of $(1-x)^{10}$, for $i=0,1,\ldots,10$.

  3. Expand $(2-x)^{-1}$ and figure out the $x^n$ and $x^{n-1}$ terms.

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  1. $(1+x^2)\left({100 \choose 1}x+\cdots+{100 \choose 3}x^3\right) \Rightarrow $ coefficient of $x^3 = {100 \choose 1}+{100 \choose 3} = 100\left(1+33 \times 49\right)= 100 \times 1618= 161800$

Follow suggestion given by Arturo. The one shown above is just to clear things for you.

Do similarly approach the other answers

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