If ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + ? + C n x n , then C 0 C 1 + C 1 C 2 + ? + C n - 1 C n is

Question

If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} C_{1} + C_{1} C_{2} + \dots + C_{n - 1} C_{n}$ is

Smart Achievers · Accepted Answer

(1)

${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ ...(i)

and ${(1 + \frac{1}{x})}^{n} = C_{0} + C_{1} \frac{1}{x} + C_{2} \frac{1}{x^{2}} + \dots + C_{n - 1} \frac{1}{x^{n - 1}} + C_{n} \frac{1}{x^{n}}$ ...(ii)

Multiplying (i) and (ii) and equating the coefficient of $x$ in $\frac{1}{x^{n}} {(1 + x)}^{2 n},$

we get $C_{0} C_{1} + C_{1} C_{2} + C_{2} C_{3} + \dots + C_{n - 1} C_{n} = {}^{2 n}C_{n + 1}$

$= \frac{(2 n)!}{(2 n - n - 1)! (n + 1)!} = \frac{(2 n)!}{(n - 1)! (n + 1)!}$

Solution

Q.
If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} C_{1} + C_{1} C_{2} + \dots + C_{n - 1} C_{n}$ is

1 $\frac{(2 n)!}{(n - 1)! (n + 1)!}$

2 $\frac{n!}{(n - 1)! (n + 1)!}$

3 $\frac{(n - 1)!}{n! (n + 1)!}$

4 none of these

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Solution

Q. If (1+x)n=C0+C1x+C2x2+⋯+Cnxn, then C0C1+C1C2+⋯+Cn-1Cn is

1 (2n)!(n-1)!(n+1)!

2 n!(n-1)!(n+1)!

3 (n-1)!n!(n+1)!