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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Binomial Theorem

If ∑_ r=0 ^5 ^ 11 C_ 2r+1 2 r+2 = m n , gcd(m, n)=1 , then m-n is...

If ∑r=0511C2r+12r+2=mn,gcd⁡(m,n)=1\sum_{r=0}^5 \frac{{ }^{11} C_{2r+1}}{2 r+2}=\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1, then m−n\mathrm{m}-\mathrm{n} is equal to _______

Correct answer

0

Step-by-step solution

(1+x)11=11C0+11C1x+11C2x2+⋯+11C11x11∫01(1+x)11dx=∫01(11C0+11C1x+11C2x2+⋯+11C11xx11)dx(1−x)1212∣01=11C0x+11C1x22+11C2x3+⋯+11C11x1212]01212−112=C0+C12+C23+⋯+C1112…(1)\begin{aligned} & (1+x)^{11}={ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11} \\ & \int_0^1(1+x)^{11} d x=\int_0^1\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x x^{11}\right) d x \\ & \left.\left.\frac{(1-x)^{12}}{12}\right|_0 ^1={ }^{11} C_{0 x}+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]_0^1 \\ & \frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots+\frac{C_{11}}{12} \ldots(1) \end{aligned} Now, ∫−10(1+x)11dx=∫−10(11C0+11C1x+11C2x2+⋯+11C11x11)dx(1+x)1212]−10=11C0x+11C1x22+11C2x33+⋯+11C11x1212]−10\begin{aligned} & \int_{-1}^0(1+x)^{11} d x=\int_{-1}^0\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11}\right) d x \\ & \left.\left.\frac{(1+x)^{12}}{12}\right]_{-1}^0={ }^{11} C_0 x+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x^3}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]_{-1}^0 \end{aligned}\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3} \cdots (1)−(2)=212−212=2[C12+C34+⋯ ]⇒∑r=05C2r+12r+2=211−112=204712=mn=2047−12=2035\begin{aligned} & (1)-(2) \\ & =\frac{2^{12}-2}{12}=2\left[\frac{C_1}{2}+\frac{C_3}{4}+\cdots\right] \\ & \Rightarrow \sum_{r=0}^5 \frac{C_{2 r+1}}{2 r+2}=\frac{2^{11}-1}{12}=\frac{2047}{12}=\frac{m}{n} \\ & =2047-12=2035 \end{aligned}$
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