Question

If \[ (1+x)^n=C_0+C_1x+C_2x^2+\cdots+C_nx^n \] where \[ C_i={}^{n}C_i, \] then \[ (C_0+C_1)(C_1+C_2)\cdots(C_{n-1}+C_n) \] equals

• A. \(C_1C_2\cdots C_n\dfrac{(n+1)^n}{n!}\)
• B. \(C_1C_2\cdots C_n\dfrac{(n-1)^2}{n}\)
• C. \(\dfrac{(2n)!}{(n!)^2}\)
• D. \(n\cdot2^{\,n-1}\)

Hint:

Remember: \[ {}^nC_r+{}^nC_{r+1} = {}^{n+1}C_{r+1} \] This identity frequently appears in product-type binomial coefficient questions.

Answer 1

The Correct Option is A

Concept: Using the identity \[ {}^nC_r+{}^nC_{r+1} = {}^{\,n+1}C_{r+1} \] each factor can be simplified.

Step 1: Transform every factor. \[\begin{aligned} C_0+C_1 &= {}^{n+1}C_1 \\ C_1+C_2 &= {}^{n+1}C_2 \\ &\vdots \\ C_{n-1}+C_n &= {}^{n+1}C_n \end{aligned}\] Hence, \[\begin{aligned} P &= \prod_{r=1}^{n} {}^{n+1}C_r \end{aligned}\]

Step 2: Express in factorial form. \[\begin{aligned} P &= \prod_{r=1}^{n} \frac{(n+1)!} {r!(n+1-r)!} \end{aligned}\] Using \[ {}^{n+1}C_r = \frac{n+1}{r} \,{}^nC_{r-1} \] we obtain \[\begin{aligned} P = C_1C_2\cdots C_n \frac{(n+1)^n}{n!} \end{aligned}\] \[\begin{aligned} \boxed{ C_1C_2\cdots C_n \frac{(n+1)^n}{n!} } \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.