The value of 1 11 C 0 + 2 11 C 1 + 3 11 C 2 + …… . + 12 11 C 11 will be

Using n C k = k n n − 1 C k − 1 For 0 ≤ k ≤ 11 k + 1 11 C k = 12 12 C k + 1 So, given expression is ⇒ 12 1 ∑ k = 0 11 12 C k + 1 ⇒ 12 1 [ ∑ k = 0 11 12 C k − 12 C 0 ] ⇒ 12 1 ( 2 12 − 1 )

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Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremProperties of Binomial coefficientsMedium2 minQB

MathematicsMediumsingle choice

The value of $\frac{^{11} C _{0}}{1} + \frac{^{11} C _{1}}{2} + \frac{^{11} C _{2}}{3} + \dots\dots . + \frac{^{11} C _{11}}{12}$ will be

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Using $^{n} C_{k} = \frac{n}{k}^{n - 1} C_{k - 1}$ For $0 \leq k \leq 11$ $\frac{^{11} C _{k}}{k + 1} = \frac{^{12} C _{k + 1}}{12}$

So, given expression is $\Rightarrow \frac{1}{12} \sum_{k = 0}^{11}^{12} C_{k + 1} \Rightarrow \frac{1}{12} [\sum_{k = 0}^{11}^{12} C_{k} -^{12} C_{0}] \Rightarrow \frac{1}{12} (2^{12} - 1)$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Properties of Binomial coefficients

⏱ 2mℹ️ Source: QB

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