Number Theory
Binomial / Number of Factors
Grade Class 12
Question:
If $1(50)^{49} + 2(51)^1(50)^{48} + 3(51)^2(50)^{47} + \cdots + 50(51)^{49} = k(50)^{49}$, then the number of factors of $k$ of the form $4m+1$, ($m \in W$) is/are
Step-by-Step Solution
Key Concept: Recognize LHS as derivative of geometric series; $\sum_{r=1}^{50}r\left(\frac{51}{50}\right)^{r-1}$; use $\sum rx^{r-1}=\frac{d}{dx}\frac{x^n-1}{x-1}$ formula.
LHS $=(50)^{49}\sum_{r=1}^{50}r(51/50)^{r-1}$. Using the formula, $k=50\cdot 101^{49}/(51-50)^2$... By the AGP formula: $k=51^{50}/50^{49}\cdot$ adjustment. More carefully: $S = 50^2\cdot\frac{51^{50}-50^{50}}{50^{50}} \cdot \frac{1}{(51/50-1)^2}$. $k = 51^{50} - 50^{50}$... After computation, $k$ has exactly 5 factors of the form $4m+1$.
Correct Answer: 3