**22.** If \[ \sum_{r=0}^{5} \frac{{^{11}C_{2r+1}}}{2r + 2} = \frac{m}{n}, \quad \gcd(m, n) = 1, \] then \( m - n \) is equal to _____.

Core Ideas Needed

1. Binomial Coefficient:
   $^{n}\!C_k = \frac{n!}{k!(n-k)!}$ gives the number of ways to choose $k$ items from $n$.

2. Symmetry:
   $^{n}\!C_k = ^{n}\!C_{n-k}$ helps pair terms like $^{11}\!C_1$ with $^{11}\!C_{10}$.

3. Evaluating a Finite Sum: Sometimes there is a closed-form identity, but for small $n$ you can safely compute each term because there are only a few.

4. Reducing Fractions: After addition, convert every term to a common denominator, add the numerators, then simplify.

Logical Walk-Through a Student Would Follow

1. List the six $k$-values from the pattern $2r+1$ with $r=0$ to $5$ → $k=1,3,5,7,9,11$.
2. Compute each binomial coefficient $^{11}\!C_k$ for those $k$'s.
3. Divide each coefficient by $(2r+2)$, i.e. $^{11}\!C_1/2,\;^{11}\!C_3/4,\;\dots,^{11}\!C_{11}/12$
4. Add the six fractions. A neat trick is to notice that the first and fifth terms are equal, the second and fourth terms are equal, and the middle one is a whole number.
5. Find a common denominator (12 is a good choice, because all denominators divide 12).
6. Sum numerators, reduce if possible, giving $\frac{m}{n}$.
7. Check gcd$(m,n)=1$, then compute $m-n$.

Why each step matters

• Listing $k$ ensures we do not miss any term.
• Symmetry cuts mental effort: once you know $^{11}\!C_3$, you also know $^{11}\!C_8$ (though here we actually need $k=7$, but the idea is the same).
• A common denominator avoids mistakes when adding fractions.
• Reducing and checking gcd guarantees the fraction is in its simplest form before subtracting.

What is the question?

We have to add together six fractions. Each fraction uses one of the binomial numbers from the row "11 choose k" and divides it by a small whole number. After adding, we reduce the answer to the simplest fraction $\frac{m}{n}$ and then subtract $n$ from $m$.

How to think about it (kid-style)

1. Binomial numbers are just the ways you can choose things. For example, $^{11}\!C_3$ tells how many ways to choose 3 marbles from 11.
2. We are only choosing the odd numbers of marbles: 1, 3, 5, 7, 9, and 11.
3. After writing those six big numbers, we divide each by 2, 4, 6, 8, 10, and 12 (notice the pattern +2 each time).
4. Finally, we add them up, change the big mixed number into a neat fraction, and do $m-n$.

So it is just a fancy way of saying: “Add six funny-looking fractions, make them tidy, and subtract.”

Step-by-Step Calculation

1. Identify the six terms ($r=0\text{ to }5$):

   r=0: &\; 2r+1 = 1, &&\; \frac{^{11}\!C_1}{2} = \frac{11}{2}\\[4pt] r=1: &\; 2r+1 = 3, &&\; \frac{^{11}\!C_3}{4} = \frac{165}{4}\\[4pt] r=2: &\; 2r+1 = 5, &&\; \frac{^{11}\!C_5}{6} = \frac{462}{6}=77\\[4pt] r=3: &\; 2r+1 = 7, &&\; \frac{^{11}\!C_7}{8} = \frac{330}{8}=\frac{165}{4}\\[4pt] r=4: &\; 2r+1 = 9, &&\; \frac{^{11}\!C_9}{10} = \frac{55}{10}=\frac{11}{2}\\[4pt] r=5: &\; 2r+1 = 11, &&\; \frac{^{11}\!C_{11}}{12} = \frac{1}{12} \end{aligned}$$

2. Express every term with denominator 12 (LCM of 2,4,6,8,10,12):

   \frac{11}{2} &= \frac{66}{12}\\ \frac{165}{4} &= \frac{495}{12}\\ 77 &= \frac{924}{12}\\ \frac{165}{4} &= \frac{495}{12}\\ \frac{11}{2} &= \frac{66}{12}\\ \frac{1}{12} &= \frac{1}{12} \end{aligned}$$

3. Add the numerators: $66+495+924+495+66+1 = 2047$ So the sum is $\frac{2047}{12}.$

4. Check for simplification: $2047 = 23\times89$, which shares no common factor with $12=2^2\times3$. Hence gcd$(2047,12)=1$.

5. Find $m-n$:
   $m = 2047, \quad n = 12 \Rightarrow m-n = 2047-12 = 2035.$

[\boxed{2035}]