If alpha, beta are natural numbers such that | vedclass

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(B) Let the sum be $S = \sum_{k=0}^{99} (100-k)(100+k)$.$S = \sum_{k=0}^{99} (100^2 - k^2) = \sum_{k=0}^{99} 100^2 - \sum_{k=0}^{99} k^2$.Since there

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$550$

Vedclass · Answer

(B) Let the sum be $S = \sum_{k=0}^{99} (100-k)(100+k)$.$S = \sum_{k=0}^{99} (100^2 - k^2) = \sum_{k=0}^{99} 100^2 - \sum_{k=0}^{99} k^2$.Since there are $100$ terms (from $k=0$ to $99$):$S = 100(100^2) - \frac{99(99+1)(2 	imes 99 + 1)}{6} = 100^3 - \frac{99 	imes 100 	imes 199}{6}$.$S = 1000000 - 33 	imes 50 	imes 199 = 1000000 - 328350 = 671650$.Given $100^{\alpha} - 199\beta = 671650$.If $\alpha = 3$,then $100^3 - 199\beta = 671650 \implies 1000000 - 671650 = 199\beta$.$328350 = 199\beta \implies \beta = \frac{328350}{199} = 1650$.Thus,the point is $(3, 1650)$.The slope of the line passing through $(3, 1650)$ and $(0, 0)$ is $m = \frac{1650 - 0}{3 - 0} = 550$.

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 the origin is:

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