coefficient of $x^2$, in $(1+x+x^2)^{10}$

Question

How to find coefficient of $x^2$, in $(1+x+x^2)^{10}$, without actually expanding it?

I think the fact $\dfrac{1-x^3}{1-x}=1+x+x^2$ may help. But can't use it!

Answer 1

Method $\#1:$

Following your process, $$(1+x+x^2)^{10}=\left(\frac{1-x^3}{1-x}\right)^{10} =(1-x^3)^{10}(1-x)^{-10}$$

$$=\left[1-\binom{10}1x^3+\binom{10}2x^6-\cdots\right]\left[1-10(-x)+\frac{-10(-10-1)}2(-x)^2+\cdots\right]$$

Clearly the coefficient of $x^2$ is $\displaystyle \frac{-10(-10-1)}2=55$

Method $\#2:$

For a clearer version of user$125053$'s answer

$$(1+x+x^2)^{10}=\left[(1+x)+x^2\right]^{10}$$ $$=\binom{10}0(1+x)^{10}+\binom{10}1(1+x)^9(x^2)^1+\binom{10}1(1+x)^8(x^2)^2\cdots$$

So, the required coefficient of $x^2$ is

the coefficient of $x^2$ in $\displaystyle\binom{10}0(1+x)^{10}+$ the coefficient of $x^0$ in $\displaystyle\binom{10}1(1+x)^9$

$\displaystyle=\binom{10}0\cdot\binom{10}2+\binom{10}1\cdot\binom90 =55 $ again

Method $\#3:$

$$(1+x+x^2)^{10}=\left[1+x(1+x)\right]^{10}$$ $$=1+\binom{10}1x(1+x)+\binom{10}2\{x(1+x)\}^2+\binom{10}3\{x(1+x)\}^3+\cdots$$ $$=1+\binom{10}1(x+x^2)+\binom{10}2\{x^2(1+x)^2\}+\binom{10}3\{x(1+x)\}^3+\cdots$$

So, the required coefficient of $x^2$ is $\displaystyle\binom{10}1+\binom{10}2=55$

Method $\#4:$

Using Multinomial Theorem, the general term of the expansion of $\displaystyle(1+x+x^2)^{10}$

is $\displaystyle\frac{10!}{a!b!c!}(1)^a(x)^b(x^2)^c=\frac{10!}{a!b!c!}x^{b+2c}$ where $a,b,c\ge0$ and $a+b+c=10$

Now we need $b+2c=2\iff2c=2-b\le2$ as $b\ge0$

$\displaystyle\implies b\le1\implies b=0,1$

If $b=0,\cdots$

If $b=1,\cdots$

Answer 2

So think about all the different ways you can get $x^2$ when you do expand it. Either you have a single $x^2$ being multiplied by $9$ other $1$'s, or $x$ multiplied my another $x$ multiplied by $8$ other $1$'s. It shouldn't be too hard to count all the ways.

Answer 3

Using the binomial theorem,

$(1 + x + x^2)^{10} = \displaystyle \sum_{j=0}^{10} \binom{10}{j} (1 + x)^j (x^2)^{10 - j}$

Using the binomial theorem again,

$\begin{align*} (1 + x + x^2)^{10} & = \displaystyle \sum_{j=0}^{10} \binom{10}{j} \left( \sum_{k=0}^j \binom{j}{k} x^k \right) (x^2)^{10 - j} \\ & = \sum_{j=0}^{10} \sum_{k=0}^j \binom{10}{j} \binom{j}{k} x^{20 + k - 2j} \end{align*}$

Now, by parity, $k$ must be even. There are only two possbilities: $k = 0, j = 9$ and $k = 2, j = 10$. Thus, the coefficient is

$\displaystyle \binom{10}{9} \binom{9}{0} + \binom{10}{10} \binom{10}{2} = 10 \cdot 1 + 1 \cdot 45 = 55$

Answer 4

As you noted: $$ [x^2]\left(1+x+x^2\right)^{10} = [x^2] \frac{(1-x^3)^{10}}{(1-x)^{10}} $$ Because $x^3$ will only contribute to terms $x^3$ and higher order $$ [x^2] \frac{(1-x^3)^{10}}{(1-x)^{10}} = [x^2] \frac{1}{(1-x)^{10}} = [x^2] \frac{1}{(1-x)^{10}} $$ Now using $n \cdot [x^n] g(x) = [x^{n-1}] g^\prime(x)$ with $n=2$ and $g(x) = \frac{1}{9}(1-x)^{-9}$: $$ [x^2] \frac{1}{(1-x)^{10}} = 3 [x^{3}] \frac{1}{9 (1-x)^9} = 3 \cdot 4 [x^4] \frac{1}{9 \cdot 8 (1-x)^8} = \frac{(11)!}{2} \cdot [x^{11}] \frac{1}{9! \cdot (1-x)} $$ Using $[x^n] \left(1-x\right)^{-1} = 1$ for $n\geq 1$ we arrive at: $$ [x^2] \frac{1}{(1-x)^{10}} = \frac{(11)!}{2 \cdot 9!} = \frac{11 \cdot 10}{2} = 55 $$

Answer 5

Since the exponent $2$ sought does not exceed the highest one in $1+x+x^2$, the answer will be the same as the coefficient of $x^2$ in $(1+x+x^2+x^3+\cdots)^{10}=(\frac1{1-x})^{10}$ which is $(-1)^2\binom{-10}2=\binom{11}2=55.$

The combinatorial interpretation is that a contribution to $x^2$ comes from choosing two of the factors $1+x+x^2$, where order is ignored but the two factors may be the same one (in which case it is the term $x^2$ that contributes rather than $x$). That this can be done in $\binom{11}2$ ways can be seen for instance by a stars-and-bars argument.

For other coefficients in $(1+x+x^2)^{10}$ such a simple argument will not work though.