The value of ∑ r = 1 10 r · C r n C r - 1 n is equal to

We have, ∑ r = 1 10 r · C r n C r - 1 n = ∑ r = 1 10 n - r + 1 ⇒ ∑ r = 1 10 r · C r n C r - 1 n = ∑ r = 1 10 n + 1 - r ⇒ ∑ r = 1 10 r · C r n C r - 1 n = 10 n + 1 - ∑ r = 1 10 r ⇒ ∑ r = 1 10 r · C r n C r - 1 n = 10 n + 1 - 55 = 10 n - 45 = 5 2 n - 9

Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremGeneralMedium2 minPYQ_2017

MathematicsMediumsingle choice

The value of $\sum_{r = 1}^{10} r \cdot \frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}}$ is equal to

Options:

Answer:

Solution:

We have,

$\sum_{r = 1}^{10} r \cdot \frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = \sum_{r = 1}^{10} (n - r + 1)$

$\Rightarrow \sum_{r = 1}^{10} r \cdot \frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = \sum_{r = 1}^{10} \{(n + 1) - r\}$

$\Rightarrow \sum_{r = 1}^{10} r \cdot \frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = 10 (n + 1) - \sum_{r = 1}^{10} r$

$\Rightarrow \sum_{r = 1}^{10} r \cdot \frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = 10 (n + 1) - 55$

$= 10 n - 45 = 5 (2 n - 9)$

Stream:BITSATSubject:MathematicsTopic:Binomial TheoremSubtopic:General

⏱ 2mℹ️ Source: PYQ_2017

Doubts & Discussion

Loading discussions...