The sum of algebraic distances from the points $(n, n^{2})$ where $n = 1, 2, 3 . . . . . .,$ $2013,$ to a straight line is zero and if such a line always passes through the point $(α, β)$ and $\frac{1}{α} (β - \frac{1007}{3}) = a b c d$ , then the value of $b + d =__________$ ( $a b c d$ is four digited number)

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answer is 5.

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Detailed Solution

$Given, \frac{a (\sum_{n = 1}^{2013} n) + b (\sum_{n = 1}^{2013} n^{2}) + c \sum_{n = 1}^{2 = 13} (1)}{\sqrt{a^{2} + b^{2}}} = 0$

$a (\frac{2013 \times 2014}{2})$

$+ b (\frac{2013 \times 2014 \times 4027}{6}) + 2013 c = 0$

$\Rightarrow a (1007) + b (\frac{1007 \times 4027}{3}) + c = 0$

$\Rightarrow α = 1007, β = \frac{1007 \times 4027}{3}$

$Now, \frac{1}{α} (β - \frac{1007}{3}) = \frac{1}{1007} (\frac{1007}{3} (4026))$

$= 1342 = a b c d$

$\Rightarrow b = 3, d = 2$

$\Rightarrow b + d = 3 + 2 = 5$

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