JEE Main 2021 (Online) 18th March Morning ShiftQuestion 1
If $\alpha,\beta$ are natural numbers such that $100^{\alpha}-199\beta=(100)(100)+(99)(101)+(98)(102)+\ldots+(1)(199)$, then the slope of the line passing through $(\alpha,\beta)$ and origin is :
Sequence and Series · Standard Sums (Σ formulas)
Solution
Formulas used. (1) $\sum_{k=1}^{100}k=5050$, $\ \sum_{k=1}^{100}k^2=338350$Step 1 - General term. The $k$-th product (from the end) is $k(200-k)$, $k=1,\dots,100$. Step 2 - Sum. $$\sum_{k=1}^{100}k(200-k)=200(5050)-338350=671650 .$$ Step 3 - Find $\alpha,\beta$. $100^\alpha-199\beta=671650$ with $\beta$ natural needs $100^\alpha>671650$, so $\alpha\ge3$. With $\alpha=3$: $$199\beta=10^6-671650=328350\ \Longrightarrow\ \beta=1650 .$$ Step 4 - Slope. $$\frac\beta\alpha=\frac{1650}{3}=550 .$$ Answer: (D)