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 \ \ \ \ \

Mathematics · Binomial Theorem

JEE Main 2025 — 22 January, Morning Shift — Question 22

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

Answer: 2035

Numerical answer — enter this value.

Step-by-step solution

\sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2} = \sum_{\substack{k=0 \\ k \ \mathrm{odd}}}^{11} \frac{\binom{11}{k}}{k+1}.

= \sum_{\substack{k=0 \\ k \ \mathrm{odd}}}^{11} \binom{11}{k} \int_{0}^{1} x^{k} \, dx = \int_{0}^{1} \sum_{\substack{k=0 \\ k \ \mathrm{odd}}}^{11} \binom{11}{k} x^{k} \, dx.

\sum_{\substack{k=0 \\ k \ \mathrm{odd}}}^{11} \binom{11}{k} x^{k} = \frac{(1+x)^{11} - (1-x)^{11}}{2}.

\therefore \quad \sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{1}{2} \int_{0}^{1} \Big( (1+x)^{11} - (1-x)^{11} \Big) \, dx.

= \frac{1}{2} \left[ \frac{(1+x)^{12}}{12} + \frac{(1-x)^{12}}{12} \right]_{0}^{1}.

= \frac{1}{24} \big( 2^{12} - 2 \big) = \frac{2047}{12}.

\frac{m}{n} = \frac{2047}{12}, \quad \gcd(m,n)=1.

\therefore \ m-n = 2047 - 12 = \boxed{2035}.

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Applications of Binomial Theorem