Binomial Theorem Question 20
Question 20 - 01 February - Shift 1
The value of
$\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots .+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !}$ is
(1) $\frac{2^{50}}{50 !}$
(2) $\frac{2^{50}}{51 !}$
(3) $\frac{2^{51}}{51 !}$
(4) $\frac{2^{51}}{50 !}$
Show Answer
Answer: (2)
Solution:
Formula: Important Results on Binomial Coefficients (iv)
$ \begin{aligned} & \sum _{r=1}^{26} \frac{1}{(2 r-1) !(51-(2 r-1)) !}=\sum _{r=1}^{26}{ }^{51} C _{(2 r-1)} \frac{1}{51 !} \\ & \quad=\frac{1}{51 !}{{ }^{51} C_1+{ }^{51} C_3+\ldots .+{ }^{51} C _{51}}=\frac{1}{51 !}(2^{50}) \end{aligned} $
