Let (1+x)n=C0+C1x+C2x2+⋯+CnxnStatement-1:∑r=0n Crsin⁡(rx)cos⁡(n−r)x=2nsin⁡(nx)Statement-2: ∑r=0n Cr=2n

# Let $\left(1+x{\right)}^{n}={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+\cdots +{C}_{n}{x}^{n}$Statement-1:$\sum _{r=0}^{n} {C}_{r}\mathrm{sin}\left(rx\right)\mathrm{cos}\left(n-r\right)x={2}^{n}\mathrm{sin}\left(nx\right)$Statement-2: $\sum _{r=0}^{n} {C}_{r}={2}^{n}$

1. A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

2. B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

3. C

STATEMENT-1 is True, STATEMENT-2 is False

4. D

STATEMENT-1 is False, STATEMENT-2 is True

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### Solution:

Let $S=\sum ^{n} {C}_{r}\mathrm{sin}\left(rx\right)\mathrm{cos}\left[\left(n-r\right)x\right]$              (1)

Using ${C}_{r}={C}_{n-r}$ , we can write

$S=\sum _{r=0}^{n} {C}_{n-r}\mathrm{sin}\left(rx\right)\mathrm{cos}\left[\left(n-r\right)x\right]$

$=\sum _{r=0}^{n} {C}_{r}\mathrm{sin}\left[\left(n-r\right)x\right]\mathrm{cos}\left(rx\right)$                   (2)

Adding (1) and (2), we get

$\begin{array}{l}2S=\sum _{r=0}^{n} {C}_{r}\left\{\mathrm{sin}\left(rx\right)\mathrm{cos}\left[\left(n-r\right)x\right]+\\ \mathrm{sin}\left[\left(n-r\right)x\right]\mathrm{cos}\left(rx\right)\right\}\\ =\sum _{r=0}^{n} {C}_{r}\mathrm{sin}\left(nx\right)=\mathrm{sin}\left(nx\right)\sum _{r=0}^{n} {C}_{r}\\ ={2}^{n}\mathrm{sin}\left(nx\right)\end{array}$

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