∑ r = 1 n ∑ p = 0 r − 1 n C r . r C p . 2 p is equal to

∑ r = 1 n n C r ( ( 1 + 2 ) r − 2 r ) = ∑ r = 1 n n C r 3 r − ∑ r = 1 n n C r 2 r = ( 4 n − 1 ) − ( 3 n − 1 ) = 4 n − 3 n

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Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremProperties of Binomial coefficientsMedium2 minQB

MathematicsMediumsingle choice

$\sum_{r = 1}^{n} \sum_{p = 0}^{r - 1}^{n} C_{r} .^{r} C_{p} . 2^{p}$ is equal to

Options:

Answer:

Solution:

$\sum_{r = 1}^{n}^{n} C_{r} ((1 + 2)^{r} - 2^{r}) = \sum_{r = 1}^{n}^{n} C_{r} 3^{r} - \sum_{r = 1}^{n}^{n} C_{r} 2^{r}$ $= (4^{n} - 1) - (3^{n} - 1) = 4^{n} - 3^{n}$

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

⏱ 2mℹ️ Source: QB

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