If 1- x n - C 0- C 1 x - C 2 x 2- C n x n- and C 0- C 1 C 1- C 2 - C n -1- C n- k - C 1 C 2 C 3- C nthen k is equal to A- nn-n-1 -B- n-1n-n-1 -C- None of theseD- n-1n-n -

The correct option is D (n+1)^nn !We have, C_0+C_1=^n+1 C_1, C_1+C_2=^n+1 C_2, …, C_n 1+C_n=^n+1 C_n ∴ The given expression =^n+1 C_1^n+1 C_2…^n+1 C_n ⋯ (1) ...

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Byju's Answer

Standard XII

Mathematics

Binomial Expression

If 1+ x n = C...

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , and $(C_{0} + C_{1}) (C_{1} + C_{2}) \dots (C_{n - 1} + C_{n}) = k \cdot C_{1} C_{2} C_{3} \dots C_{n}$
then $k$ is equal to

\frac{(n + 1)^{n}}{n!}

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\frac{n^{n}}{(n - 1)!}

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\frac{(n + 1)^{n}}{(n - 1)!}

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None of these

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Solution

The correct option is A $\frac{(n + 1)^{n}}{n!}$
We have,
$C_{0} + C_{1} =^{n + 1} C_{1}, C_{1} + C_{2} =^{n + 1} C_{2}, \dots,$
$C_{n - 1} + C_{n} =^{n + 1} C_{n}$
$∴ The given expression =^{n + 1} C_{1}^{n + 1} C_{2} \dots^{n + 1} C_{n} \dots (1)$
$Now,^{n + 1} C_{r} = \frac{(n + 1)!}{r! (n - r + 1)!} = \frac{(n + 1) n!}{r! (n - r + 1) (n - r)!} \Rightarrow^{n + 1} C_{r} = \frac{n + 1}{n - r + 1}^{n} C_{r} \dots (2)$

Putting $n = 1, 2, 3, \dots, n$ in Eq. $(2)$ , we get
$(C_{0} + C_{1}) (C_{1} + C_{2}) \dots (C_{n - 1} + C_{n}) = \frac{n + 1}{n} C_{1} \cdot \frac{n + 1}{n - 1} C_{2} \dots \frac{n + 1}{1} C_{n} = \frac{(n + 1)^{n}}{n!} C_{1} C_{2} C_{3} \dots C_{n}$

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If (1+x)n=C0+C1x+C2x2+⋯+Cnxn, and (C0+C1)(C1+C2)⋯(Cn−1+Cn)=k⋅C1C2C3⋯Cn then k is equal to

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then $k$ is equal to