Question

$\frac{69}{420} + \frac{69 \cdot 70}{420 \cdot 421} + \frac{69 \cdot 70 \cdot 71}{420 \cdot 421 \cdot 422} + \dots = \frac{m}{n}$,

where $m, n$ are relatively prime positive integers. Find $100m + n$.

Answer 1

7250

Answer 2

The given series is identified as a specific form of a hypergeometric series $_2F_1(a,1;b;1)$, excluding its first term (for $k=0$). The sum of $_2F_1(a,1;b;1)$ is evaluated using Gauss's hypergeometric theorem, which states $_2F_1(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$. By setting $a=69$, $b=1$, $c=420$, the sum from $k=0$ to $\infty$ is calculated as $\frac{419}{350}$. Since the original series starts from $k=1$, we subtract the $k=0$ term (which is 1) from this sum. This yields $S = \frac{419}{350} - 1 = \frac{69}{350}$. Comparing this with $\frac{m}{n}$, we get $m=69$ and $n=350$. These are verified to be relatively prime. Finally, $100m+n$ is computed as $100(69) + 350 = 7250$.