For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.

Question

For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.

I tried to use Newton's binomial formula as: $$ (1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right) $$ and that's all I can think of, other then just to compute it.

Answer 1

Apply the trinomial theorem: $$(1+x+x^2)^6=\sum_{a+b+c=6,\:0\le a,b,c\le 6} \binom{6}{a,b,c}(1)^a(x)^b(x^2)^c$$ To get the $x^6$ term, the possibilities are $(a,b,c)=(3,0,3),(2,2,2),(1,4,1),(0,6,0)$. Hence the $x^6$ term is $$\binom{6}{3,0,3}x^6+\binom{6}{2,2,2}x^6+\binom{6}{1,4,1}x^6+\binom{6}{0,6,0}x^6=141x^6$$

Answer 2

You can use recursion:

$P(k)=(1+x+x^2)^k$

$[x^6]P(k)=[x^6]P(k-1)+[x^5]P(k-1)+[x^4]P(k-1)$

where $[x^n]P$ is the coefficient of $x^n$ in $P$.

Answer 3

Here's a solution sketch: One method would be to use $x^2+x+1=\frac {x^3-1}{x-1}$. Then you can write the expression as $(x^3-1)^6 (x-1)^{-6}$. Expand, separately, $(x^3-1)^6$ and $(x-1)^{-6}$ using the binomial theorem, and see which terms can together make up $x^6$. There would be only $3$ such terms.

Answer 4

When you expand the product you get $3^6$ summands, each summand corresponds to whethere $1,x$ or $x^2$ was picked in each of the factors.

Notice that $6$ is the average "degree", so in order to get $x^6$ you need to select the same number of $1$ as $x^2$, and therefore you need to select an even number of $x$. So if we select $x$ $2i$ times we select $1$ $3-i$ times and $x^2$ also $3-i$ times. The number of ways this can be dome is $\binom{6}{2i,3-i,3-i}$.

We can get that the number of ways to do this is $\sum\limits_{i=0}^3 \binom{6}{2i,3-i,3-i}$.

Hence the answer is $\binom{6}{0,3,3} + \binom{6}{2,2,2} + \binom{6}{4,1,1} + \binom{6}{6,0,0} = 20 + 90 + 30 + 1 = 141$

Answer 5

When expanding the product $(x^0 + x^1 + x^2)^6$, each term looks like $x^{k_1} x^{k_2} \cdots x^{k_6} = x^{k_1+k_2+\dots+k_6}$ where all $k_i\in \{0,1,2\}$. Hence, the coefficient of $x^6$ in the expansion is the number of six-tuples $(k_1,k_2,\dots,k_6)$ such that $k_1+k_2+\dots+k_6=6$ and all $k_i\in\{0,1,2\}$.

The number of $2$s in such a partition can be $0$, $1$, $2$ or $3$ and for each there is exactly one possible partition up to permuation: \begin{align} 6 &= 1+1+1+1+1+1 \qquad\text{1 permutation}, \\ &= 1+1+1+1+2+0 \qquad\text{$\frac{6!}{4!\,1!\,1!}=30$ permutations}, \\ &= 1+1+2+2+0+0 \qquad\text{$\frac{6!}{2!\,2!\,2!}=90$ permutations}, \\ &= 2+2+2+0+0+0 \qquad\text{$\frac{6!}{3!\,3!}=20$ permutations.} \end{align}

Hence, in total there are $141$ summands equal to $x^6$ in the expansion, so the term you are looking for is $141 x^6$.

Answer 6

Hint:

You can use the multinomial formula: $$(a+b+c)^6=\sum_{\substack{i+j+k=6\\ 0\le i,j, k\le 6}}\frac{6}{i!\,j!\,k!}\,a^i b^j c^k.$$ Settin $a=1$, $b=x$, $c=x^2$, you have to determine the solutions in natural numbers of the linear system \begin{cases} i+j+k=6 ,\\ j+2k=6. \end{cases}

Answer 7

This question might be easier to solve by hand.

However, the general way to solve it is to know that:

$$\frac1{(1-x)^k}=\sum_{n=0}^\infty \binom{n+k-1}{k-1}x^k$$

Then use:

$$x^2+x+1=\frac{1-x^3}{1-x}$$

So:

$$(x^2+x+1)^6 =(1-x^3)^6\frac1{(1-x)^3}$$

So the coefficient of $x^n$ is $$\sum_{k=0}^6(-1)^k\binom 6k\binom{n-3k+5}5$$

This gives, for $n=6,$

$$\binom{11}5-\binom61\binom85 +\binom62\binom55$$

because the other terms are zero.

Answer 8

$$(1 + x+ x^2)^6 = \frac{(1-x^3)^6}{(1-x)^6} = (1-x^3)^6 (1-x)^{-6}$$ We wish to calculate the coefficient of $x^6$. I shall denote $I = (1-x^3)^6$ and $J =(1-x)^{-6}$. We want the coefficient of $x^6$ in $IJ$, denoted as $[IJ]_{6}$. In general, $[P]_k$ denotes the coefficient of $x^k$ in $P$.

Note that $$(1+x)^{-n} = \sum_{k=0}^{\infty} {-n\choose k}x^k = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} x^k$$ so, $$(1-x)^{-6} = \sum_{k=0}^\infty {-6 \choose k} (-x)^k$$ Also, $$[IJ]_6 = [I]_0 [J]_6 + [I]_1 [J]_5 + [I]_2 [J]_4 + [I]_3 [J]_3 + [I]_4 [J]_2 + [I]_5 [J]_1 + [I]_6 [J]_0$$ and we know that for $0\le k\le 6$, $$[J]_k = {-6\choose k}(-1)^k = {k+5\choose k}$$ and using the binomial theorem $$[I]_k = \begin{cases}0 & k \ne 0,3,6 \\ 1 & k =0 \\ -6 & k = 3 \\ 15 & k =6\end{cases}$$ I'll leave the calculations to you.

Answer 9

We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance \begin{align*} [x^k](1+x)^n=\binom{n}{k}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{[x^6]}&\color{blue}{((1+x)+x^2)^6}\\ &=[x^6]\sum_{k=0}^6\binom{6}{k}x^{2k}(1+x)^{6-k}\tag{2}\\ &=\sum_{k=0}^3\binom{6}{k}[x^{6-2k}](1+x)^{6k-k}\tag{3}\\ &=\sum_{k=0}^3\binom{6}{k}\binom{6-k}{6-2k}\tag{4}\\ &=\binom{6}{0}\binom{6}{6}+\binom{6}{1}\binom{5}{4}+\binom{6}{2}\binom{4}{2}+\binom{6}{3}\binom{3}{0}\\ &=1\cdot1+6\cdot 5+15\cdot 6+20\cdot 1\\ &\,\,\color{blue}{=141} \end{align*}

Comment:

• In (2) we use (1) and apply the binomial theorem.

• In (3) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$. We also set the upper limit to $3$ since other indices do not contribute.

• In (4) we select the coefficient of $x^{6-2k}$.