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If (1+x)^n=\sum_{r=0}^n C_r x^r \text { then }\left(1+\frac{C_1}{C_0}\right)\left(1+\frac{C_2}{C_1}\right) \ldots \ldots\left(1+\frac{C_n}{C_{n-1}}\right)=

Option: 1

\frac{n^{n-1}}{(n-1) !}


Option: 2

\frac{n^{n-1}}{(n-1) !}


Option: 3

\frac{(n+1)^{n-1}}{(n-1) !}


Option: 4

\frac{(n+1)^{n+1}}{n !}


Answers (1)

best_answer

We have \left(1+\frac{C_1}{C_0}\right)\left(1+\frac{C_2}{C_1}\right) \ldots \ldots\left(1+\frac{C_n}{C_{n-1}}\right)=\left(1+\frac{n}{1}\right)\left(1+\frac{n(n-1) / 2 !}{n}\right) \ldots \ldots\left(1+\frac{1}{n}\right)

=\left(\frac{1+n}{1}\right)\left(\frac{1+n}{2}\right)\left(\frac{1+n}{3}\right) \ldots \ldots\left(\frac{1+n}{n}\right)=\frac{(n+1)^n}{n !}

Trick : Put n=1,2,3 \ldots \ldots, S_1=1+\frac{{ }^1 C_1}{{ }^1 C_0}=2, S_2=\left(1+\frac{{ }^2 C_1}{{ }^2 C_1}\right)\left(1+\frac{{ }^2 C_2}{{ }^2 C_1}\right)=\frac{9}{2}

Which is given by option (c) n=1, \quad \frac{(1+1)^1}{1 !}=2 ; \text { For } n=2, \quad \frac{(2+1)^2}{2 !}=\frac{9}{2}

Posted by

himanshu.meshram

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