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Using the binomial theorem find the coefficient of $x^2$ in the expression $(1+x+x^2)^{10}$.

Please help me by explaining how to proceed with this question.

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closed as off-topic by Grigory M, Claude Leibovici, Sami Ben Romdhane, Najib Idrissi, Daniel Rust Apr 20 at 13:14

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You could use the binomial theorem twice. –  user88595 Apr 20 at 8:50

4 Answers 4

To get an $x^2$, we need to pick one factor in the product with $x^2$ and the rest $1$, $$\overbrace{(\color{#C00000}{1}+x+x^2)\cdots}^{n-1}(1+x+\color{#C00000}{x^2}) $$ that is, $\binom{10}{1}$, or two factors with $x^1$ and the rest $1$, $$\overbrace{(\color{#C00000}{1}+x+x^2)\cdots}^{n-2}(1+\color{#C00000}{x}+x^2)(1+\color{#C00000}{x}+x^2)$$ that is, $\binom{10}{2}$. Thus, the coefficient of $x^2$ is $$ \binom{10}{1}+\binom{10}{2}=55 $$


Another Approach $$ \begin{align} (1+x+x^2)^{10} &=\sum_{k=0}^{10}\binom{10}{k}(x+x^2)^k\\ \end{align} $$ the only terms in the sum that contribute an $x^2$ are those with $k=1$ and $k=2$: $$ \begin{align} &\binom{10}{1}(x+x^2)+\binom{10}{2}(x+x^2)^2\\ &=\binom{10}{1}x+\left[\binom{10}{1}+\binom{10}{2}\right]x^2+2\binom{10}{2}x^3+\binom{10}{2}x^4 \end{align} $$ giving the coefficient of $x^2$ to be $$ \binom{10}{1}+\binom{10}{2}=55 $$

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You typo'd 100 for 10 :) –  Anant Apr 20 at 9:57
    
@Anant: ack! I misread :-) I have fixed it. Thanks. –  robjohn Apr 20 at 13:36

Hint: Consider $(a+b)^{10}$, where $a=(1+x)$ and $b=x^2$.

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I tried to do so but I could only expand the expression. I couldn't find the coefficient of x^2. –  Shivangi puri Apr 20 at 8:34
    
Once expanded, you need to pick out the terms that will contain $x^2$, and expand again for each of them. If you prefer, you could think about whether taking $a=1$, or $a=x$ (and corresponding $b$) will make this any more straightforward. –  Joshua Pepper Apr 20 at 8:38
    
Thank you. I was able to solve it. –  Shivangi puri Apr 20 at 8:42

Another variation: You could solve it somewhat mechanically: Let $[x^k]$ denote the coefficient of $x^k$ in a polynomial $p(x)=\sum_{j=0}^{n}a_{j}x^{j}$, i.e. $[x^k]p(x)=a_k$.

We have to calculate $$[x^2]\left(1+x+x^2\right)^{10}$$

Therefore

\begin{align*} [x^2]&\left(1+x+x^2\right)^{10}\\ &=[x^2]\left(1+\left(x+x^2\right)\right)^{10}\\ &=[x^2]\sum_{j=0}^{10}\binom{10}{j}\left(x+x^2\right)^j\\ &=[x^2]\sum_{j=0}^{10}\binom{10}{j}x^j\left(1+x\right)^j\\ &=\sum_{{j=0}\atop{2-j\geq0}}^{10}\binom{10}{j}[x^{2-j}]\left(1+x\right)^j\\ &=\sum_{j=0}^{2}\binom{10}{j}[x^{2-j}]\sum_{l=0}^{j}\binom{j}{l}x^l\\ &=\sum_{j=0}^{2}\binom{10}{j}\binom{j}{2-j}\\ &=\binom{10}{1}\binom{1}{1}+\binom{10}{2}\binom{2}{0}\\ &=55\end{align*}

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hint: $\displaystyle(1+x+x^2)^{10} = \bigg(x+\frac{1}{2}+\frac{i\sqrt{3}}{2}\bigg)^{10}\bigg(x+\frac{1}{2}+\frac{i\sqrt{3}}{2}\bigg)^{10}$

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