Binomial Theorem
AGP — Coefficient of x^48 in Weighted Sum
nta_pyq_2026_jan
Grade 11
Question:
The coefficient of $x^{48}$ in $(1+x)+2(1+x)^2+3(1+x)^3+\cdots+100(1+x)^{100}$ is equal to
$100\cdot{}^{100}C_{49}-{}^{100}C_{48}$
${}^{100}C_{50}+{}^{101}C_{49}$
$100\cdot{}^{101}C_{49}-{}^{101}C_{50}$
$100\cdot{}^{100}C_{49}-{}^{100}C_{50}$
Step-by-Step Solution
Key Concept: Let $r=1+x$. $S=\sum_{k=1}^{100}k\cdot r^k$. Use AGP: $S(1-r)=-100r^{101}+\sum_{k=1}^{100}r^k$. So $S=\frac{-(1+x)^{101}}{x^2}+\frac{1}{x^2}+\frac{100(1+x)^{101}}{x}$.
Coeff of $x^{48}=100\cdot{}^{101}C_{49}-{}^{101}C_{50}$.
Correct Answer: 3