Question:
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 :
Correct Option: , 2
Solution:
$S=(100)(100)+(99)(101)+(98)(102) \ldots . .$
$\ldots(2)(198)+(1)(199)$
$S=\sum_{x=0}^{99}(100-x)(100+x)=\sum 100^{2}-x^{2}$
$=100^{3}-\frac{99 \times 100 \times 199}{6}$
$\alpha=3$
$\beta=1650$
slope $=\frac{1650}{3}=550$
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