If 1+frac{cos theta}{2}+frac{cos 2 theta}{4}+ | vedclass

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(B) The given series is $S = 1 + \frac{\cos 	heta}{2} + \frac{\cos 2 	heta}{4} + \frac{\cos 3 	heta}{8} + \ldots$ This is an infinite series of the

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(B) The given series is $S = 1 + \frac{\cos 	heta}{2} + \frac{\cos 2 	heta}{4} + \frac{\cos 3 	heta}{8} + \ldots$ This is an infinite series of the form $\sum_{n=0}^{\infty} \frac{\cos(n	heta)}{2^n}$. Using the formula $\sum_{n=0}^{\infty} r^n \cos(n	heta) = \frac{1-r \cos 	heta}{1-2r \cos 	heta + r^2}$ with $r = \frac{1}{2}$: $S = \frac{1-\frac{1}{2} \cos 	heta}{1-2(\frac{1}{2}) \cos 	heta + (\frac{1}{2})^2} = \frac{1-\frac{1}{2} \cos 	heta}{1-\cos 	heta + \frac{1}{4}} = \frac{\frac{2-\cos 	heta}{2}}{\frac{5-4 \cos 	heta}{4}} = \frac{2(2-\cos 	heta)}{5-4 \cos 	heta} = \frac{4-2 \cos 	heta}{5-4 \cos 	heta}$. Comparing this with $\frac{a-2 \cos 	heta}{5+b \cos 	heta}$,we get $a = 4$ and $b = -4$. Therefore,$(a-b)^2 = (4 - (-4))^2 = (8)^2 = 64$.

If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots = \frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in R$,then $(a-b)^2=$

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