Question

Sum of the coefficients of $x^4$ and $x^6$ in the expansion of $(1 + x - x^2)^6$ is

• A. 121
• B. -91
• C. 11
• D. 31

Hint:

For multinomial expansions like $(a + bx + cx^2)^n$, solve $b + 2c = k$ for the coefficient of $x^k$, using $\frac{n!}{a!b!c!} (-1)^c$ where $a + b + c = n$.

Answer 1

The Correct Option is C

Use the multinomial theorem for $(1 + x - x^2)^6 = \sum_{a+b+c=6} \frac{6!}{a!b!c!} 1^a x^b (-x^2)^c = \sum_{a+b+c=6} \frac{6!}{a!b!c!} (-1)^c x^{b+2c}$.
We need coefficients of $x^4$ and $x^6$.
For $x^4$: Solve $b + 2c = 4$, $a + b + c = 6$. Possible triples $(a, b, c)$:
- $c = 0$, $b = 4$, $a = 2$: $\frac{6!}{2!4!0!} (-1)^0 = \frac{720}{2 \cdot 24} = 15$
- $c = 1$, $b = 2$, $a = 3$: $\frac{6!}{3!2!1!} (-1)^1 = \frac{720}{6 \cdot 2 \cdot 1} \cdot (-1) = -60$
- $c = 2$, $b = 0$, $a = 4$: $\frac{6!}{4!0!2!} (-1)^2 = \frac{720}{24 \cdot 2} = 15$
Coefficient: $15 - 60 + 15 = -30$.
For $x^6$: Solve $b + 2c = 6$, $a + b + c = 6$. Possible triples:
- $c = 0$, $b = 6$, $a = 0$: $\frac{6!}{0!6!0!} (-1)^0 = 1$
- $c = 1$, $b = 4$, $a = 1$: $\frac{6!}{1!4!1!} (-1)^1 = \frac{720}{1 \cdot 24 \cdot 1} \cdot (-1) = -30$
- $c = 2$, $b = 2$, $a = 2$: $\frac{6!}{2!2!2!} (-1)^2 = \frac{720}{2 \cdot 2 \cdot 2} = 90$
- $c = 3$, $b = 0$, $a = 3$: $\frac{6!}{3!0!3!} (-1)^3 = \frac{720}{6 \cdot 6} \cdot (-1) = -20$
Coefficient: $1 - 30 + 90 - 20 = 41$.
Sum: $-30 + 41 = 11$. Option (3) is correct. Options (1), (2), and (4) do not match.