If (1+x)n=Co+C1x+C2x2+...+Cnxn. Then, CoC1+C1C2+...+Cn−1Cn is equal to
2n!(n−1)!(n+1)!
(2n−1)!(n−1)!(n+1)!
(2n)!(n+2)!(n+1)!
None of these
To find CoC1+C1C2+...+Cn−1Cn:
Step 1: Find (x+1)n
Given that (1+x)n=Co+C1x+C2x2+...+Cnxn. Then we have to find CoC1+C1C2+...+Cn–1Cn
(x+1)n=Coxn+C1xn−1+C2xn−2+...+Cn
Step 2: Multiply (1+x)n and (x+1)n
(1+x)2n=(Co+C1x+C2x2+...+Cnxn)(Coxn+C1xn−1+C2xn−2+...+Cn)
Step 3: Compare the coefficients of xn−1
2nn+1=CoC1+C1C2+...+Cn−1Cn
2n!(n−1)!(n+1)! =CoC1+C1C2+...+Cn−1Cn
Hence, Option (A): 2n!(n−1)!(n+1)! is the correct option.