Let 1 - xn - -r - 0n - Crxr and C1C0 - 2C2C1 - 3C8C2 - - - nCnCn - 1 - 1k nn - 1-Then- the value of k is

The correct option is D 2 Given, C_1C_0 + 2C_2C_1 + 3C_3C_2 + .... + nC_nC_n 1 = 1kn(n + 1) ∴n1 + 2·n(n 1)2n + 3·n(n 1)(n 2)3!×2!n(n 1) ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

Let 1 + xn ...

Question

Let $(1 + x)^{n} = n \sum r = 0 = C_{r} x^{r}$ and $\frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{8}}{C_{2}} + . . . + \frac{n C_{n}}{C_{n - 1}} = \frac{1}{k} n (n + 1)$ .
Then, the value of $k$ is

\frac{1}{2}

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2

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\frac{1}{3}

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3

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Solution

The correct option is D $2$
Given,
$\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + . . . . + n \frac{C_{n}}{C_{n - 1}} = \frac{1}{k} n (n + 1)$

$∴ \frac{n}{1} + 2 \cdot \frac{n (n - 1)}{2 n} + 3 \cdot \frac{n (n - 1) (n - 2)}{3!} \times \frac{2!}{n (n - 1)} + . . . . + n \cdot \frac{1}{n} = \frac{1}{k} n (n + 1)$
$\Rightarrow n + (n - 1) + (n - 2) + . . . . + 1 = \frac{1}{k} n (n + 1)$
$\Rightarrow \frac{n (n + 1)}{2} = \frac{1}{k} n (n + 1)$

$\Rightarrow k = 2$

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

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Let (1+x)n=n∑r=0=Crxr and C1C0+2C2C1+3C8C2+...+nCnCn−1=1kn(n+1).Then, the value of k is

The correct option is D 2Given,C1C0+2C2C1+3C3C2+....+nCnCn−1=1kn(n+1)∴n1+2⋅n(n−1)2n+3⋅n(n−1)(n−2)3!×2!n(n−1)+....+n⋅1n=1kn(n+1)⇒n+(n−1)+(n−2)+....+1=1kn(n+1)⇒n(n+1)2=1kn(n+1)⇒k=2

Let $(1 + x)^{n} = n \sum r = 0 = C_{r} x^{r}$ and $\frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{8}}{C_{2}} + . . . + \frac{n C_{n}}{C_{n - 1}} = \frac{1}{k} n (n + 1)$ .
Then, the value of $k$ is