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7.Binomial Theorem
hard
If $\sum_{ r =0}^5 \frac{{ }^{11} C _{2 r +1}}{2 r +2}=\frac{ m }{ n }, \operatorname{gcd}( m , n )=1$, then $m - n$ is equal to _____
A$2785$
B$2035$
C$5039$
D$2235$
(JEE MAIN-2025)
Solution
$\int_0^1(1+x)^{11} d x=\left[C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots\right]_0^1$
$\frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\frac{C_3}{4}+\ldots$
$\int_{-1}^0(1+x)^{11} d x=\left[C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots\right]_{-1}^0$
$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\ldots$
$\frac{2^{12}-2}{12}=2\left(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots\right)$
$\frac{C_1}{2}+\frac{C_3}{4}-\frac{C_5}{6}+\ldots=\frac{2^{11}-1}{12}=\frac{2047}{12}$
$\frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\frac{C_3}{4}+\ldots$
$\int_{-1}^0(1+x)^{11} d x=\left[C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots\right]_{-1}^0$
$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\ldots$
$\frac{2^{12}-2}{12}=2\left(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots\right)$
$\frac{C_1}{2}+\frac{C_3}{4}-\frac{C_5}{6}+\ldots=\frac{2^{11}-1}{12}=\frac{2047}{12}$
Standard 11
Mathematics
