Question

The coefficient of $x^7$ in the expression $\left(1+x\right)^{10}+x\left(1+x\right)^{9}+x^{2}\left(1+x\right)^{8}+...+x^{10}$ is :

• A. $420$
• B. $330$
• C. $210$
• D. $120$

Answer 1

The Correct Option is B

To find the coefficient of $x^7$ in the expression $(1+x)^{10} + x(1+x)^9 + x^2(1+x)^8 + \ldots + x^{10}$, we need to evaluate the series step by step.

The given expression is a sum: $(1+x)^{10} + x(1+x)^9 + x^2(1+x)^8 + \ldots + x^{10}$. Each term in this series contributes to the coefficient of $x^7$ in different ways. Let's consider each term separately:

1. In the first term, $(1+x)^{10}$, use the binomial theorem to find the coefficient of $x^7$. The coefficient of $x^7$ is $\binom{10}{7}$. $\binom{10}{7} = 120$.
2. In the second term, $x(1+x)^9$, consider the coefficient of $x^6$ in $(1+x)^9$. Multiply it by $x$ to get $x^7$. The coefficient of $x^6$ is $\binom{9}{6}$. $\binom{9}{6} = 84$.
3. For the third term, $x^2(1+x)^8$, consider the coefficient of $x^5$ in $(1+x)^8$. Multiply it by $x^2$. The coefficient of $x^5$ is $\binom{8}{5}$. $\binom{8}{5} = 56$.
4. Continuing this process, we look for the coefficients of terms such that the sum of the power of $x$ in any term plus the power of $x$ of the exterior coefficient gives $7$.
   • $x^3(1+x)^7\Rightarrow \binom{7}{4}=35$.
   • $x^4(1+x)^6\Rightarrow \binom{6}{3}=20$.
   • $x^5(1+x)^5\Rightarrow \binom{5}{2}=10$.
   • $x^6(1+x)^4\Rightarrow \binom{4}{1}=4$.
   • $x^7(1+x)^3\Rightarrow \binom{3}{0}=1$.

Summing these coefficients gives us: $120 + 84 + 56 + 35 + 20 + 10 + 4 + 1 = 330$.

Therefore, the coefficient of $x^7$ in the expression is $330$.