Question

The Coefficient of $x^7$ in $(1-x+2x^3)^{10}$ is ______

Answer 1

120

Answer 2

To find the coefficient of $x^7$ in the expansion of $(1-x+2x^3)^{10}$, we use the multinomial theorem.

The general term in the expansion of $(a+b+c)^n$ is given by: $T_{p,q,r} = \frac{n!}{p!q!r!} a^p b^q c^r$ where $p+q+r=n$ and $p,q,r$ are non-negative integers.

In our case, $a=1$, $b=-x$, $c=2x^3$, and $n=10$. So, the general term is: $T_{p,q,r} = \frac{10!}{p!q!r!} (1)^p (-x)^q (2x^3)^r$ $T_{p,q,r} = \frac{10!}{p!q!r!} (1)^p (-1)^q (x)^q (2)^r (x^3)^r$ $T_{p,q,r} = \frac{10!}{p!q!r!} (-1)^q (2)^r x^{q+3r}$

We need the coefficient of $x^7$, so we set the exponent of $x$ to 7: $q+3r = 7$ Also, the sum of the powers must be equal to $n$: $p+q+r = 10$ We need to find all possible non-negative integer solutions for $(p,q,r)$ that satisfy these two equations.

Let's find the possible values for $r$:

1. Case 1: $r=0$ Substituting $r=0$ into $q+3r=7$: $q+3(0) = 7 \Rightarrow q=7$. Now substitute $q=7, r=0$ into $p+q+r=10$: $p+7+0 = 10 \Rightarrow p=3$. So, $(p,q,r) = (3,7,0)$. The coefficient for this term is $\frac{10!}{3!7!0!} (-1)^7 (2)^0$. $\frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times (-1) \times 1 = 120 \times (-1) = -120$.

2. Case 2: $r=1$ Substituting $r=1$ into $q+3r=7$: $q+3(1) = 7 \Rightarrow q=4$. Now substitute $q=4, r=1$ into $p+q+r=10$: $p+4+1 = 10 \Rightarrow p=5$. So, $(p,q,r) = (5,4,1)$. The coefficient for this term is $\frac{10!}{5!4!1!} (-1)^4 (2)^1$. $\frac{10 \times 9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1 \times 1} \times (1) \times (2) = (10 \times 3 \times 7 \times 2) \times 2 = 840 \times 2 = 1680$.

3. Case 3: $r=2$ Substituting $r=2$ into $q+3r=7$: $q+3(2) = 7 \Rightarrow q+6=7 \Rightarrow q=1$. Now substitute $q=1, r=2$ into $p+q+r=10$: $p+1+2 = 10 \Rightarrow p=7$. So, $(p,q,r) = (7,1,2)$. The coefficient for this term is $\frac{10!}{7!1!2!} (-1)^1 (2)^2$. $\frac{10 \times 9 \times 8}{2 \times 1} \times (-1) \times 4 = 360 \times (-1) \times 4 = -1440$.

4. Case 4: $r=3$ Substituting $r=3$ into $q+3r=7$: $q+3(3) = 7 \Rightarrow q+9=7 \Rightarrow q=-2$. This is not possible, as $q$ must be a non-negative integer. So we stop here.

The total coefficient of $x^7$ is the sum of the coefficients from all valid cases: Total Coefficient = (Coefficient from Case 1) + (Coefficient from Case 2) + (Coefficient from Case 3) Total Coefficient = $-120 + 1680 - 1440$ Total Coefficient = $1560 - 1440$ Total Coefficient = $120$