\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\end{aligned}$

$$\frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots+\frac{C_{11}}{12} \cdots$$ ........(1)

Now,

$$ \begin{aligned} & \int_{-1}^0(1+x)^{11} d x=\int_{-1}^0\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11}\right) d x \\ & \left.\left.\frac{(1+x)^{12}}{12}\right]_{-1}^0={ }^{11} C_0 x+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x^3}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]_{-1}^0 \end{aligned} $$

$$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3} \cdots$$ …….(2)

$\begin{aligned} & \text { (1) }-(2) \\\\ & =\frac{2^{12}-2}{12}=2\left[\frac{C_1}{2}+\frac{C_3}{4}+\cdots\right] \\\\ & \Rightarrow \sum_{r=0}^5 \frac{C_{2 r+1}}{2 r+2}=\frac{2^{11}-1}{12}=\frac{2047}{12}=\frac{m}{n} \\\\ & =2047-12=2035\end{aligned}$

19:T4d5,

$\begin{aligned} & (1+x)^{11}={ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11} \\ & \int_0^1(1+x)^{11} d x=\int_0^1\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x{ }^2+\cdots+{ }^{11} C_{11} x{ }^{11}\right) d x \\ & \left.\left.\frac{(1-x)^{12}}{12}\right]_0^1={ }^{11} C_{0 x}+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]_0^1\end{aligned}$

$$\frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots+\frac{C_{11}}{12} \cdots$$ ........(1)

अब,

$$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3} \cdots$$ …….(2)

$\begin{aligned} & \text { (1) }-(2) \\\\ & =\frac{2^{12}-2}{12}=2\left[

Binomial Theorem

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