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

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Q. $(C_0 + C_1)(C_1 + C_2) ... (C_{n-1} + C_n)$ is equal to

Binomial Theorem

Solution:

$\left(C_{0}+C_{1}\right)\left(C_{1}+C_{2}\right) \ldots \left(C_{n-1}+C_{n}\right)$

$= C_{0}\left(1+\frac{C_{1}}{C_{0}}\right)+C_{1}\left(1+\frac{C_{2}}{C_{1}}\right)\ldots C_{n-1}\left(1+\frac{C_{n}}{C_{n-1}}\right)$

$=\left\{C_{0} C_{1} C_{2} ... C_{n-1}\right\}\left(1+\frac{C_{1}}{C_{0}}\right)\left(1+\frac{C_{2}}{C_{1}}\right)\ldots \left(1+\frac{C_{n}}{C_{n-1}}\right)$

$= \left\{C_{0} C_{1} C_{2} ... C_{n-1}\right\} \left(n+1\right)\left(1+\frac{n-1}{2}\right)\ldots\left(1+\frac{1}{n}\right)$

$=\left\{C_{0} C_{1} C_{2} ... C_{n-1}\right\} \left(n+1\right)\left(\frac{n+1}{2}\right)\left(\frac{n+1}{3}\right)\ldots\left(\frac{n+1}{n}\right)$

$= \frac{\left\{C_{0} C_{1} C_{2} ... C_{n-1}\right\} \left(n+1\right)^{n}}{n!}$