Question

If $\sum_{r=0}^{5}\frac{^{11}C_{2r}}{2r+2} = \frac{m}{n}$, gcd(m, n) = 1, then m – n is equal to ___.

Answer 1

1721

Answer 2

The given sum is $\sum_{r=0}^{5}\frac{^{11}C_{2r}}{2r+2}$. We expand the sum by substituting values of r from 0 to 5:

1. For $r=0$: $\frac{^{11}C_0}{2} = \frac{1}{2}$
2. For $r=1$: $\frac{^{11}C_2}{4} = \frac{55}{4}$
3. For $r=2$: $\frac{^{11}C_4}{6} = \frac{330}{6} = 55$
4. For $r=3$: $\frac{^{11}C_6}{8} = \frac{462}{8} = \frac{231}{4}$
5. For $r=4$: $\frac{^{11}C_8}{10} = \frac{165}{10} = \frac{33}{2}$
6. For $r=5$: $\frac{^{11}C_{10}}{12} = \frac{11}{12}$

Summing these terms: $S = \frac{1}{2} + \frac{55}{4} + 55 + \frac{231}{4} + \frac{33}{2} + \frac{11}{12}$

To combine these fractions, we find a common denominator, which is 12: $S = \frac{6}{12} + \frac{165}{12} + \frac{660}{12} + \frac{693}{12} + \frac{198}{12} + \frac{11}{12}$ $S = \frac{6 + 165 + 660 + 693 + 198 + 11}{12} = \frac{1733}{12}$

Given that $S = \frac{m}{n}$ and gcd(m, n) = 1, we have $m=1733$ and $n=12$. Since 1733 is not divisible by 2 or 3, and 12 is $2^2 \times 3$, gcd(1733, 12) = 1. Finally, we calculate $m-n$: $m-n = 1733 - 12 = 1721$.