If $\sum_{\mathrm{r}=0}^{5} \frac{{ }^{11} \mathrm{C}_{2 \mathrm{r}+1}}{2 \mathrm{r}+2}=\frac{\mathrm{m}}{\mathrm{n}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}-\mathrm{n}$ is equal to ____.

Binomial Theorem — JEE Main · 2025 · 22 Jan 2025 · Shift 1 Question with Solution

\sum_{r = 0}^{5} \frac{^{11} C _{2 r + 1}}{2 r + 2} = \frac{m}{n}

, where

gcd (m, n) = 1

, then

m - n

is equal to ____.

Answer: 2035

We need

r = 0 \sum 5 \frac{^{11} C _{2 r + 1}}{2 r + 2} = \frac{C _{1}}{2} + \frac{C _{3}}{4} + \frac{C _{5}}{6} + \frac{C _{7}}{8} + \frac{C _{9}}{10} + \frac{C _{11}}{12}

, the sum of

\frac{C _{k}}{k + 1}

over odd

k

from 1 to 11.

Define

F (x) = k = 0 \sum 11 \frac{C _{k} x ^{k + 1}}{k + 1}

, which is the antiderivative of

(1 + x)^{11} = \sum C_{k} x^{k}

vanishing at

x = 0

F (1) = \int_{0}^{1} (1 + x)^{11} d x = [\frac{( 1 + x ) ^{12}}{12}]_{0}^{1} = \frac{2 ^{12} - 1}{12} = \frac{4095}{12} .

F (1) - F (- 1) = \int_{- 1}^{1} (1 + x)^{11} d x = [\frac{( 1 + x ) ^{12}}{12}]_{- 1}^{1} = \frac{2 ^{12} - 0}{12} = \frac{4096}{12} .

F (- 1) = F (1) - \frac{4096}{12} = \frac{4095 - 4096}{12} = - \frac{1}{12}

.

Since

F (x) = \sum_{k} \frac{C _{k} x ^{k + 1}}{k + 1}

, the term for index

k

contributes

x^{k + 1}

; at

x = - 1

this is

(- 1)^{k + 1}

, which is

+ 1

for odd

k

and

- 1

for even

k

. So

F (1) + F (- 1) = k odd \sum \frac{2 C _{k}}{k + 1} = 2 (\frac{C _{1}}{2} + \frac{C _{3}}{4} + \frac{C _{5}}{6} + \frac{C _{7}}{8} + \frac{C _{9}}{10} + \frac{C _{11}}{12}) .

F (1) + F (- 1) = \frac{4095}{12} - \frac{1}{12} = \frac{4094}{12} = \frac{2047}{6} .

\frac{C _{1}}{2} + \frac{C _{3}}{4} + \dots + \frac{C _{11}}{12} = \frac{1}{2} \cdot \frac{2047}{6} = \frac{2047}{12} .

Since

2047 = 23 \times 89

and

12 = 2^{2} \times 3

share no common factor,

m = 2047, n = 12

, and

m - n = 2047 - 12 = 2035.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Integer answer