For z in mathbb{C},if (1+z)^n = 1 + { }^n C_1 | vedclass

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(C) Given the binomial expansion $(1+z)^n = \sum_{r=0}^{n} { }^n C_r z^r$. Let $z = \cos x + i \sin x = e^{ix}$. Then $(1 + e^{ix})^n = \sum_{r=0}^{n}

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) Given the binomial expansion $(1+z)^n = \sum_{r=0}^{n} { }^n C_r z^r$. Let $z = \cos x + i \sin x = e^{ix}$. Then $(1 + e^{ix})^n = \sum_{r=0}^{n} { }^n C_r e^{irx}$. Using $1 + e^{ix} = 1 + \cos x + i \sin x = 2 \cos^2 \frac{x}{2} + 2i \sin \frac{x}{2} \cos \frac{x}{2} = 2 \cos \frac{x}{2} (\cos \frac{x}{2} + i \sin \frac{x}{2}) = 2 \cos \frac{x}{2} e^{i \frac{x}{2}}$. So,$(1 + e^{ix})^n = (2 \cos \frac{x}{2})^n e^{i \frac{nx}{2}} = (2 \cos \frac{x}{2})^n (\cos \frac{nx}{2} + i \sin \frac{nx}{2})$. Equating the imaginary parts of $\sum_{r=0}^{n} { }^n C_r e^{irx} = \sum_{r=0}^{n} { }^n C_r (\cos rx + i \sin rx)$,we get $\sum_{r=0}^{n} { }^n C_r \sin(rx) = (2 \cos \frac{x}{2})^n \sin(\frac{nx}{2})$. For $n = 100$,$\sum_{r=0}^{100} { }^{100} C_r \sin(rx) = (2 \cos \frac{x}{2})^{100} \sin(\frac{100x}{2}) = (2 \cos \frac{x}{2})^{100} \sin(50x)$. Comparing this with the given expression $(2 \cos \frac{x}{2})^{100} \sin(kx)$,we find $k = 50$.

For $z \in \mathbb{C}$,if $(1+z)^n = 1 + { }^n C_1 z + { }^n C_2 z^2 + \ldots + { }^n C_n z^n$ and $\sum_{r=0}^{100} { }^{100} C_r \sin(rx) = \left(2 \cos \frac{x}{2}\right)^{100} \sin(kx)$,then $k =$

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