Binomial Theorem
Coefficient in Multi-factor Expansion
nta_pyq_2024_jan
Grade 11
Question:
In the expansion of $(1+x)(1-x^2)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5$, $x\neq 0$, the sum of the coefficient of $x^3$ and $x^{-13}$ is equal to ______.
Step-by-Step Solution
Key Concept: Recognize $\left(1+\frac{1}{x}\right)^3 = 1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}$. So the expression equals $\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} = \frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}}$. Finding coefficients of $x^3$ and $x^{-13}$ reduces to finding coefficients of $x^{18}$ and $x^2$ in $(1+x)^{17}-x(1+x)^{17}$.
$(1+x)(1-x^2)\left(1+\frac{1}{x}\right)^{15}\cdot\frac{1}{x^{15}}$
$=\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}}=\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}}$.
Coeff of $x^3$: need coeff of $x^{18}$ in $(1+x)^{17}-x(1+x)^{17}$ $= 0-1=-1$.
Coeff of $x^{-13}$: need coeff of $x^2$ in $(1+x)^{17}-x(1+x)^{17}$ $=\binom{17}{2}-\binom{17}{1}=136-17=119$.
Sum $= -1+119=118$.
Correct Answer: 118