What is the coefficient of $x^4$ in the trinomial expansion of $(1+x+x^2)^{10}$?

Question

I'm confused whether it is to be done directly or I actually have to factorise it first?

Answer 1

Method 1

General term for this trinomial expansion is $$\dfrac{10!}{a!b!c!}x^{b+2c}$$ where $a+b+c=10$ and $a$, $b$ and $c$ are non-negative integers. Now, we need $$b+2c=4$$ Hence, we can have $(a,b,c)=(8,0,2)$ or $(6,4,0)$ or $(7,2,1)$. So the required coefficient is $$\dfrac{10!}{8!0!2!}+\dfrac{10!}{6!4!0!}+\dfrac{10!}{7!2!1!}=45+210+360$$ $$=615$$

Method 2

$$(1+x+x^2)^{10}=\left(\dfrac{1-x^3}{1-x}\right)^{10}$$ $$=(1-10x^3+45x^6-\cdots)(1-x)^{-10}$$ $$=(1-10x^3+45x^6-\cdots)(1+\binom{10}{1}x+\binom{11}{2}x^2+\binom{12}{3}x^3+\binom{13}{4}x^4+\cdots)$$ Required coefficient is $\binom{13}{4}-100=615$.

Answer 2

This is how I would do it: we have, for any $a,b,$

$$(a+b)^{10}=a^{10}+10a^9b+45a^8b^2+\text{higher powers of } b$$

Put $a=x+1, b=x^2$: $$((x+1)+x^2)^{10}=(x+1)^{10}+10(x+1)^9x^2+45(x+1)^8x^4+\text{higher powers of }x$$

So now you just add up the coefficient of $x^4$ in $(x+1)^{10}$, the coefficient of $x^2$ in $10(x+1)^9$, and the constant coefficient in $45(x+1)^8$.

I'm afraid your answer of $210$ is too small.

Answer 3

There are only three ways to get a factor of $x^4$ in the expansion, let's just count each way.

I. Two factors of $x^2$, all the rest $1's$. There are $\binom {10}2=45$ ways to do that.

II. One factor of $x^2$ ($10$ choices), and two factors of $x$ ($\binom 92=36$ choices), so $360$.

III. Four factors of $x$, $\binom {10}4=210$.

So the answer is $$45+360+210=\boxed {615}$$

Answer 4

Using multinomial theorem we have: $$\left(x^{2}+x+1\right)^{10}=\sum_{k_{1}=0}^{10}\sum_{k_{2}=0}^{k_{1}}{{10}\choose{k_1}}{{k_1}\choose{k_2}}\left(x^{2}\right)^{\left(10-k_{1}\right)}\left(x\right)^{k_{2}}$$

Then we want the non-negative solutions to the following equation:

$$2\color{red}{k_{1}}-\color{blue}{k_{2}}=16$$

using this we have the coefficient which is :

$${{10}\choose{\color{red}{8}}}{\color{red}{8}\choose{\color{blue}0}}+{{10}\choose{\color{red}9}}{{\color{red}9}\choose{\color{blue}2}}+{{{10}}\choose{\color{red}{10}}}{{\color{red}{10}}\choose{\color{blue}4}}=\frac{10!}{8!\cdot2!}+10\cdot\frac{9!}{7!2!}+\frac{10!}{6!4!}=\color{green} {\boxed{615}}$$

Answer 5

Assuming $1-x\ne0$

$$(1+x+x^2)^{10}=(1-x^3)^{10}(1-x)^{-10}$$

So, we need the coefficient of $x^4$ in $(1-x)^{-10}$

which will be $$\dfrac{(-10)(-11)(-12)(-13)}{4!}$$ using Binomial series