Let (1+x)n=C0+C1x+C2x2+$$…$$+Cnxn$$ and $$C1C0+2C2C1+3C3C2+$$⋯$$+nCnCn$$−$$1=1kn(n+1), then value of k is

rCrCr−$$1=rn$$!r!(n$$−$$r)!(r$$−$$1)!(n$$−$$r+1)!n!⇒ ∑$$r=1n$$ rCrCr−$$1=$$∑$$r=1n$$ (n$$−$$r+1)=12n(n+1)∴ $$k=2$$.

Binomial theorem for positive integral Index

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Question

Let $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ and $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots + n \frac{C_{n}}{C_{n - 1}} = \frac{1}{k} n (n + 1),$ then
value of $k$ is

Moderate

Solution

$\begin{array}{l} r \frac{C_{r}}{C_{r - 1}} = r \frac{n!}{r! (n - r)!} \frac{(r - 1)! (n - r + 1)!}{n!} \\ \Rightarrow \sum_{r = 1}^{n} r \frac{C_{r}}{C_{r - 1}} = \sum_{r = 1}^{n} (n - r + 1) = \frac{1}{2} n (n + 1) \\ ∴ k = 2 . \end{array}$

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