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AP EAMCET · Maths · Binomial Theorem

If \(C_j={ }^n C_j\), then \(C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots+C_{n-r} C_n=\)

  1. A \(\frac{(2 n) !}{(n-2 r) !(n+2 r) !}\)
  2. B \(\frac{(2 n) !}{(n-r) !(n+r) !}\)
  3. C \(2 \mathrm{n}_{\mathrm{C}_{\mathrm{r}}}\)
  4. D \(2 n_{C_{r+1}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{(2 n) !}{(n-r) !(n+r) !}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \text {Since }(1+x)^{2 n}=(1+x)^n(x+1)^n \\ & \Rightarrow(1+x)^{2 n}=\left(C_0+C_1 x+C_2 x^2+\ldots+C_n x^n\right) \\ & \left(C_0 x^n+C_1 x^{n-1}+C_2 x^{n-2}+\ldots+C_n\right) \end{aligned}\) Now equation co-efficient of \(x^{n-r}\) on both side…