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Question Number 101231 by 175 last updated on 01/Jul/20

Commented by Dwaipayan Shikari last updated on 01/Jul/20
$(1/n).(π/6)lim_(n→∞) ((2^(1/n) −1)/(1/n)) Σ_(r=0) ^n (2^(r/n) cos(2^(r/n) (π/6))) (π/6) log2 ∫_0 ^1 2^x cos(2^x (π/6))dx ∫_0 ^1 cos(2^x (π/6))2^x (π/6)log2dx ∫_(π/6) ^(π/3) cost dt =[sint]_(π/6) ^(π/3) =(((√3)−1)/2) {suppose 2^x (π/6)=t}$
$\frac{1}{n} . \frac{π}{6} \lim_{n \to \infty} \frac{2^{\frac{1}{n}} - 1}{\frac{1}{n}} \underset{r = 0}{\sum^{n}} (2^{\frac{r}{n}} c o s (2^{\frac{r}{n}} \frac{π}{6}))$ $\frac{π}{6} l o g 2 \int_{0}^{1} 2^{x} c o s (2^{x} \frac{π}{6}) d x$ $\int_{0}^{1} c o s (2^{x} \frac{π}{6}) 2^{x} \frac{π}{6} l o g 2 d x$ $\int_{\frac{π}{6}}^{\frac{π}{3}} c o s t d t = {[s i n t]}_{\frac{π}{6}}^{\frac{π}{3}} = \frac{\sqrt{3} - 1}{2}$ ${s u p p o s e 2^{x} \frac{π}{6} = t}$
	Answered by smridha last updated on 01/Jul/20

	$\frac{π}{6} \lim_{n \to \infty} \frac{2^{\frac{1}{n}} - 1}{\frac{1}{n}} . [\lim_{n \to \infty} \frac{1}{n} . \underset{r = 0}{\sum^{n}} 2^{\frac{r}{n}} . c o s (2^{\frac{r}{n}} . \frac{π}{6})]$ $= \frac{π}{6} . l n (2) . \int_{0}^{1} 2^{x} . c o s (2^{x} . \frac{π}{6}) d x$ $= \frac{π}{6} . l n 2 . \frac{6}{π . l n (2)} \int_{0}^{1} d [s i n (2^{x} . \frac{π}{6})]$ $= 1 . {[s i n (2^{x} . \frac{π}{6})]}_{0}^{1} = [s i n \frac{π}{3} - s i n \frac{π}{6}]$ $= \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2} .$
	Commented by 175 last updated on 01/Jul/20
	thanx
	Commented by smridha last updated on 01/Jul/20
	welcome...but careful there is a group ��
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