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

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False, STATEMENT-2 is True

Solution:

Let $S = \sum^{n} C_{r} \sin (r x) \cos [(n - r) x]$ (1)

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

$S = \sum_{r = 0}^{n} C_{n - r} \sin (r x) \cos [(n - r) x]$

$= \sum_{r = 0}^{n} C_{r} \sin [(n - r) x] \cos (r x)$ (2)

Adding (1) and (2), we get

$\begin{array}{l} 2 S = \sum_{r = 0}^{n} C_{r} {\sin (r x) \cos [(n - r) x] + \\ \sin [(n - r) x] \cos (r x)} \\ = \sum_{r = 0}^{n} C_{r} \sin (n x) = \sin (n x) \sum_{r = 0}^{n} C_{r} \\ = 2^{n} \sin (n x) \end{array}$

$\Rightarrow S = 2^{n - 1} \sin (n x)$

Let $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$
Statement-1: $\sum_{r = 0}^{n} C_{r} \sin (r x) \cos (n - r) x = 2^{n} \sin (n x)$
Statement-2: $\sum_{r = 0}^{n} C_{r} = 2^{n}$

Solution:

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