C0+C1C1+C2…Cn−1+Cn is equal to - Infinity Learn by Sri Chaitanya

A
$(C_{0} C_{1} C_{2} \dots C_{n - 1}) (n + 1)$
B
$(C_{0} C_{1} C_{2} \dots C_{n - 1}) (n + 1)^{n}$
C
$\frac{(C_{0} C_{1} C_{2} \dots C_{n - 1}) (n + 1)^{n}}{n!}$
D
None of these above

Solution:

$\begin{array}{l} (C_{0} + C_{1}) (C_{1} + C_{2}) \dots (C_{n - 1 +} C_{n}) \\ = C_{0} (1 + \frac{C_{1}}{C_{0}}) C_{1} (1 + \frac{C_{2}}{C_{1}}) \dots C_{n - 1} (1 + \frac{C_{n}}{C_{n - 1}}) \\ = \{C_{0} C_{1} C_{2} \dots C_{n - 1}\} (1 + \frac{C_{1}}{C_{0}}) (1 + \frac{C_{2}}{C_{1}}) \dots (1 + \frac{C_{n}}{C_{n - 1}}) \\ = \{C_{0} C_{1} C_{2} \dots C_{n - 1}\} (n + 1) (1 + \frac{n - 1}{2}) \dots (1 + \frac{1}{n}) \\ = \{C_{0} C_{1} C_{2} \dots C_{n - 1}\} (n + 1) (\frac{n + 1}{2}) (\frac{n + 1}{3}) \dots (\frac{n + 1}{n}) \\ = \frac{\{C_{0} C_{1} C_{2} \dots C_{n - 1}\} (n + 1)^{n}}{n!} \end{array}$

$(C_{0} + C_{1}) (C_{1} + C_{2}) \dots (C_{n - 1 +} C_{n})$ is equal to

Solution:

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