Calculating the value of A

First let's generalize the equation $A = \frac{1*98 + 2*97 +...+ 97*2 + 98*1}{1*2 + 2*3 + ... + 97*98 + 98*99}$ to any given equation of the form $\frac{\sum _{ n=1 }^{ k }{ n(k+1-n) }}{\sum _{ n=1 }^{ k }{ n(n+1) } }$ where k is the value being multiplied by 1 (in this question k=98).

Then simplify using summation rules including the facts that $\sum _{ n=1 }^{ k }{ n^{2} } =\frac { (n)(n+1)(2n+1) }{ 6 } \hspace{10 mm} and \hspace{10 mm} \sum _{ n=1 }^{ k }{ n } =\frac { (n)(n+1) }{ 2 }$ .

Now the equation can be written as: $\frac{\frac{k(k+1)(k+2)}{3}}{\frac{k(k+1)(k+2)}{6}} = \frac{1}{2}$ Since the specific case k=98 can be written in this form, it follows that $\boxed{A=\frac{1}{2}}$

use sigma (summation) series to solve the question