Find the value of - r -110 r n C T - n C r -1A- 10 nB- 9 n-45C- 9 nD- 10 n -45

The correct option is D 10n 45 ^nC_r/^nC_r 1=n!/(n r)!× r!(r 1)!× (n r+1)!/n! =n r+1/r ⇒∑_r=1^10 r^nC_r/^nC_r 1=∑_r=1^10(n r+1) =∑_r=1^10(n+1) ∑_r=1^10 r =10(n ...

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Byju's Answer

Standard XI

Mathematics

Binomial Coefficients

Find the valu...

Question

Find the value of $\sum_{r = 1}^{10} r \frac{^{n} C_{r}}{^{n} C_{r - 1}}$

10n-45

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Solution

The correct option is A
10n-45

$\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n!}{(n - r)! \times r!} \frac{(r - 1)! \times (n - r + 1)!}{n!}$
= $\frac{n - r + 1}{r}$
$\Rightarrow \sum_{r = 1}^{10} r \frac{^{n} C_{r}}{^{n} C_{r - 1}} = \sum_{r = 1}^{10} (n - r + 1)$
= $\sum_{r = 1}^{10} (n + 1) - \sum_{r = 1}^{10} r$
= $10 (n + 1) - \frac{10 \times 11}{2}$
=10(n+1)-55
=10n-45

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Find the value of ∑10r=1rnCrnCr−1

The correct option is A 10n-45 nCrnCr−1=n!(n−r)!×r!(r−1)!×(n−r+1)!n! =n−r+1r ⇒∑10r=1rnCrnCr−1=∑10r=1(n−r+1) =∑10r=1(n+1)−∑10r=1r =10(n+1)−10×112 =10(n+1)-55 =10n-45

Find the value of $\sum_{r = 1}^{10} r \frac{^{n} C_{r}}{^{n} C_{r - 1}}$