(C0 + C1)(C1 + C2) ... (Cn-1 + Cn) is equal to

Question

(C0 + C1)(C1 + C2) ... (Cn-1 + Cn) is equal to (A) (C0 C1 C2 ... Cn-1 )(n+1) (B) (C0 C1 C2 ... Cn-1 )(n+1)n (C) ((C0 C1 C2 ... Cn-1 )(n+1)n/n!) (D) None of these

Tardigrade Admin · Accepted Answer

(C0+C1)(C1+C2) ldots (Cn-1+Cn) = C0(1+(C1/C0))+C1(1+(C2/C1)) ldots Cn-1(1+(Cn/Cn-1)) = C0 C1 C2 ... Cn-1 (1+(C1/C0))(1+(C2/C1)) ldots (1+(Cn/Cn-1)) = C0 C1 C2 ... Cn-1 (n+1)(1+(n-1/2)) ldots(1+(1/n)) = C0 C1 C2 ... Cn-1 (n+1)((n+1/2))((n+1/3)) ldots((n+1/n)) = ( C0 C1 C2 ... Cn-1 (n+1)n/n!)

Q. $(C_{0} + C_{1}) (C_{1} + C_{2}) ... (C_{n - 1} + C_{n})$ is equal to

$(C_{0} C_{1} C_{2} ... C_{n - 1}) (n + 1)$

$(C_{0} C_{1} C_{2} ... C_{n - 1}) (n + 1)^{n}$

$\frac{( C _{0} C _{1} C _{2} ... C _{n - 1} ) ( n + 1 ) ^{n}}{n !}$

None of these

Q. (C0​+C1​)(C1​+C2​)...(Cn−1​+Cn​) is equal to

(C0​C1​C2​...Cn−1​)(n+1)

(C0​C1​C2​...Cn−1​)(n+1)n

n!(C0​C1​C2​...Cn−1​)(n+1)n​

None of these

Q. $(C_{0} + C_{1}) (C_{1} + C_{2}) ... (C_{n - 1} + C_{n})$ is equal to

$(C_{0} C_{1} C_{2} ... C_{n - 1}) (n + 1)$

$(C_{0} C_{1} C_{2} ... C_{n - 1}) (n + 1)^{n}$

$\frac{( C _{0} C _{1} C _{2} ... C _{n - 1} ) ( n + 1 ) ^{n}}{n !}$