Question

If $\alpha $, and $\beta $ are natural numbers such that $10{0^\alpha } - 199\beta = \left( {100} \right)\left( {100} \right) + \left( {99} \right)\left( {101} \right) + \left( {98} \right)\left( {102} \right) + ....\left( 1 \right)\left( {199} \right)$. Then what is the slope of the line passing through the point $\left( {\alpha ,\beta } \right)$ and the origin?
A. $510$
B. $550$
C. $540$
D. $530$

Answer 1

Hint: First, solve the right-hand side of the given equation. Apply the formula of the sum of squares of natural numbers and find the sum. Equate the sum with the left-hand side and get the values of the $\alpha $, and $\beta $. In the end, use the formula of slope to get the required answer.

Formula Used:
$\sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
$\sum\limits_{n = 1}^n 1 = n$
The formula of slope: If a line passes through the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$, then the slope of the line is $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

Complete step by step solution:
Given: $\alpha $ and $\beta $ are natural numbers
$10{0^\alpha } - 199\beta = \left( {100} \right)\left( {100} \right) + \left( {99} \right)\left( {101} \right) + \left( {98} \right)\left( {102} \right) + ....\left( 1 \right)\left( {199} \right)$
Let’s simplify the right-hand side of the above equation.
We can rewrite the right-hand side as follows:
$10{0^\alpha } - 199\beta = \sum\limits_{k = 0}^{99} {\left( {100 - k} \right)\left( {100 + k} \right)} $
$ \Rightarrow 10{0^\alpha } - 199\beta = \sum\limits_{k = 0}^{99} {\left( {{{\left( {100} \right)}^2} - {k^2}} \right)} $ [Since $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$]
Apply the summation on every term on the right-hand side.
$10{0^\alpha } - 199\beta = \sum\limits_{k = 0}^{99} {{{\left( {100} \right)}^2} - \sum\limits_{k = 0}^{99} {{k^2}} } $
$ \Rightarrow 10{0^\alpha } - 199\beta = {\left( {100} \right)^2}\sum\limits_{k = 0}^{99} {1 - \sum\limits_{k = 0}^{99} {{k^2}} } $
Now use the formula of the sum of squares of natural numbers.
$10{0^\alpha } - 199\beta = {\left( {100} \right)^2}\left( {100} \right) - \dfrac{{99\left( {100} \right)\left( {199} \right)}}{6}$
$ \Rightarrow 10{0^\alpha } - 199\beta = {\left( {100} \right)^3} - 199\left( {1650} \right)$
Equate with the left-hand side.
$\alpha = 3$ and $\beta = 1650$

Let’s calculate the slope of the line passing through the points $\left( {3,1650} \right)$ and the origin.
Let $m$ be the slope of the line.
Apply the formula of slope.
$m = \dfrac{{0 - 1650}}{{0 - 3}}$
$ \Rightarrow m = \dfrac{{1650}}{3}$
$ \Rightarrow m = 550$
Thus, the slope of the line is $550$.

Option ‘B’ is correct

Note: Do not get confused with the expansion of the summation $\sum\limits_{k = 0}^n {{a_k}} $. The correct expansion of the summation is: $\sum\limits_{k = 0}^n {{a_k}} = {a_0} + \sum\limits_{k = 1}^n {{a_k}} $