If , gcd (m, n) = 1, then m – n is equal to __________. [2025]

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Solution

Q.
If $\sum_{r = 0}^{5} \frac{{}^{11}C_{2 r + 1}}{2 r + 2} = \frac{m}{n}$ , gcd (m, n) = 1, then m – n is equal to __________. [2025]

Ans.
(2035)

We know that,

$C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + . . . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{(n + 1)}$ ... (i)

$C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - . . . . . . + {(- 1)}^{n} \frac{C_{n}}{n + 1} = \frac{1}{n + 1}$ ... (ii)

Subtracting (ii) from (i), we get

$2 [\frac{C_{1}}{2} + \frac{C_{3}}{4} + . . . . . .] = \frac{2^{n + 1} - 1}{(n + 1)} - \frac{1}{(n + 1)}$

Put n = 11, we get

$2 [\frac{C_{1}}{2} + \frac{C_{3}}{4} + . . . . . . + \frac{C_{11}}{12}] = \frac{2^{12} - 1}{12} - \frac{1}{12}$

$\Rightarrow \frac{C_{1}}{2} + \frac{C_{3}}{4} + . . . . . . + \frac{C_{11}}{12} = \frac{1}{2} [\frac{2^{12} - 2}{12}]$ ... (iii)

Now, $\sum_{r = 0}^{5} \frac{{}^{11}C_{2 r + 1}}{2 r + 2} = \frac{m}{n}$

$\Rightarrow \frac{{}^{11}C_{1}}{2} + \frac{{}^{11}C_{3}}{4} + \frac{{}^{11}C_{5}}{6} + . . . . . . + \frac{{}^{11}C_{11}}{12} = \frac{m}{n}$

$\Rightarrow \frac{1}{2} [\frac{2^{12} - 2}{12}] = \frac{m}{n}$ [From (iii)]

$\Rightarrow \frac{2^{11} - 1}{12} = \frac{m}{n} \Rightarrow \frac{m}{n} = \frac{2047}{12}$

$∴$ m – n = 2047 – 12 = 2035.

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