C1C0-2-C2C1-3-C3C2-n-CnCn-1-

The correct option is D n(n+1)2 ^nC_r/^nC_r 1= n (r 1)/r ⇒^nC_r+1/^nC_r= n r/r+1 Tr for the given sequence is Tr = ∑_r=0^n 1(r+1)(n r/r+1) ...

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Standard XII

Mathematics

Combination

C1C0+2.C2C1+3...

Question

$\frac{C_{1}}{C_{0}} + 2. \frac{C_{2}}{C_{1}} + 3. \frac{C_{3}}{C_{2}} + . . . . . + n . \frac{C_{n}}{C_{n - 1}}$ =

\frac{n (n + 1)}{2}

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\frac{n (n - 1)}{2}

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\frac{(n - 1) (n + 1)}{2}

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\frac{n (n + 2)}{2}

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Solution

The correct option is D $\frac{n (n + 1)}{2}$
$\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - (r - 1)}{r}$
$\Rightarrow \frac{^{n} C_{r + 1}}{^{n} C_{r}} = \frac{n - r}{r + 1}$
Tr for the given sequence is
$T r = \sum_{r = 0}^{n - 1} (r + 1) (\frac{n - r}{r + 1})$
$= \sum_{r = 0}^{n - 1} n - r = \sum_{r = 0}^{n - 1} n - \sum_{r = 0}^{n - 1} r$
$= n (n + 1) - \frac{(n + 1) (n)}{2}$
$= \frac{n (n + 1)}{2}$

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Similar questions

Q. Prove that

\frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{3}}{C_{2}} + . . . . + \frac{n . C_{n}}{C_{n - 1}} = \frac{n (n + 1)}{2}

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . + C_{n} x^{n}$ , then $\frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{3}}{C_{2}} + . . . . . . . . + \frac{n C_{n}}{C_{n - 1}} =$

\frac{c_{1}}{c_{0}} + 2. \frac{c_{2}}{c_{1}} + 3. \frac{c_{3}}{c_{2}} + . . . + n . \frac{c_{n}}{n_{1}}

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . + C_{n} x^{n}$ , then

$\frac{C_{1}}{C_{0}}$ + $\frac{2 C_{2}}{C_{1}}$ + $\frac{3 C_{3}}{C_{2}}$ + ........+ $\frac{n C_{n}}{C_{n - 1}}$ =

\frac{c_{1}}{c_{0}} + 2 \frac{c_{2}}{c_{1}} + 3 \frac{c_{3}}{c_{2}} + . . . . . . . . . + n \frac{c_{n}}{c_{n - 1}} = \frac{n (n + 1)}{2}

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C1C0+2.C2C1+3.C3C2+.....+n.CnCn−1=

The correct option is D n(n+1)2nCrnCr−1=n−(r−1)r⇒nCr+1nCr=n−rr+1Tr for the given sequence isTr=∑n−1r=0(r+1)(n−rr+1)=∑n−1r=0n−r=∑n−1r=0n−∑n−1r=0r=n(n+1)−(n+1)(n)2=n(n+1)2

$\frac{C_{1}}{C_{0}} + 2. \frac{C_{2}}{C_{1}} + 3. \frac{C_{3}}{C_{2}} + . . . . . + n . \frac{C_{n}}{C_{n - 1}}$ =