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  • Find the value of <img alt="(1 \cdot 2) C_{2}+(2 \cdot 3)C_{3}+\ldots \ldots+\{(n-1) \cdot n\} C_{n}" src="http://entrancecorner.oncodecogs.com/gif.latex?%281%20%5Ccdot%202%29%20C_%7B2%7D&

Find the value of (1 \cdot 2) C_{2}+(2 \cdot 3)C_{3}+\ldots \ldots+\{(n-1) \cdot n\} C_{n}.

 If \begin{array}{l}{ \mathrm{C}_{0}, \mathrm{C}_{1}, \mathrm{C}_{2}, \ldots, \mathrm{C}_{n} }\end{array} be binomial coefficients in the expansion of (1+x)^n

Option: 1

n(n+1) 2^{n-1}


Option: 2

n(n+1) 2^{n-2}


Option: 3

n(n-1) 2^{n-1}


Option: 4

n(n-1) 2^{n-2}


Answers (1)

best_answer

As we learnt

1^2.C_{1}+2^2 \cdot C_{2}+3^2 \cdot C_{3}+---+n^2 \cdot C_{n}=\sum_{r=0}^{n} r^2 \cdot{ }^{n} C_{r}=n \cdot 2^{n-1} + n(n-1)2^{n-2}

and 

C_{1}+2 \cdot C_{2}+3 \cdot C_{3}+---+n \cdot C_{n}=\sum_{r=0}^{n} r \cdot{ }^{n} C_{r}=n \cdot 2^{n-1}

 

Now,

Given expression can be written as 

\sum_{r=0}^{n} (r-1)r\,.\, ^{n} C_{r}\\\\ = \sum_{r=0}^{n} r^2\,.\, ^{n} C_{r} - \sum_{r=0}^{n} r\,.\, ^{n} C_{r}\\\\ = n.2^{n-1} + n(n-1).2^{n-2} - n.2^{n-1}\\\\ = n(n-1).2^{n-2}\\\\

Posted by

seema garhwal

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