Question

$\displaystyle \sf \lim_{n \to \infty} \bigg( \sum \limits^{n + 9}_{k = 10}\frac{2^{11 \frac{k - 9}{n} } }{ \log_{2}e^{ \frac{n}{11} } } - \sum \limits_{k = 0}^{n - 1} \frac{58}{\pi \sqrt{(n - k)(n + k)} } \bigg )$
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Answer 1

Step-by-step explanation:

u can check this miss sizuka

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\lim_{n \to \infty} \bigg( \sum \limits^{n + 9}_{k = 10}\frac{2^{\frac{11(k - 9)}{n} } }{ \log_{2}e^{ \frac{n}{11} } } - \sum \limits_{k = 0}^{n - 1} \dfrac{58}{\pi \sqrt{(n - k)(n + k)} } \bigg )$

can be rewritten as

$\rm \:  =  \: \lim_{n \to \infty}\sum \limits^{n + 9}_{k = 10}\frac{2^{11 \frac{k - 9}{n} } }{ \log_{2}e^{ \frac{n}{11} } } - \lim_{n \to \infty}\sum \limits_{k = 0}^{n - 1} \frac{58}{\pi \sqrt{(n - k)(n + k)} }$

$\rm \:  =  \: x - y$

where,

$\rm \:  x=  \: \lim_{n \to \infty}\sum \limits^{n + 9}_{k = 10}\frac{2^{11 \frac{k - 9}{n} } }{ \log_{2}e^{ \frac{n}{11} } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \rm \: y \: = \: \lim_{n \to \infty}\sum \limits_{k = 0}^{n - 1} \frac{58}{\pi \sqrt{(n - k)(n + k)} }$

Now, Consider

$\rm :\longmapsto\: x=  \: \lim_{n \to \infty}\sum \limits^{n + 9}_{k = 10}\frac{2^{\frac{11(k - 9)}{n} } }{ \log_{2}e^{ \frac{n}{11} } }$

$\rm :\longmapsto\: x= \lim_{n \to \infty} \dfrac{11}{n} \sum \limits^{n + 9}_{k = 10}\frac{2^{\frac{11(k - 9)}{n} } }{ \log_{2}e}$

$\rm :\longmapsto\: x=  \: \lim_{n \to \infty} \dfrac{11}{n} \sum \limits^{n + 9}_{k = 10}\frac{2^{\frac{11(k - 9)}{n} } }{ \log_{2}e}$

$\rm :\longmapsto\: x=  \: \lim_{n \to \infty} \dfrac{11}{n} \sum \limits^{n}_{k = 1}\frac{2^{\frac{11k}{n} } }{ \log_{2}e}$

Now, using Limit as a sum of definite integrals, we get

$\rm :\longmapsto\:x = \dfrac{11}{ log_{2}(e) } \displaystyle\int_0^1\rm {2}^{11x} \: dx$

$\rm :\longmapsto\:x = \dfrac{11}{ log_{2}(e) } \bigg[\dfrac{ {2}^{11x} }{11 \times log_{e}(2) } \bigg]_0^1$

$\bf\implies \:x = {2}^{11} - {2}^{0} = 2048 - 1 = 2047$

Now, Consider

$\rm :\longmapsto\:y \: = \: \lim_{n \to \infty}\sum \limits_{k = 0}^{n - 1} \dfrac{58}{\pi \sqrt{(n - k)(n + k)} }$

$\rm :\longmapsto\:y \: = \: \lim_{n \to \infty}\sum \limits_{k = 0}^{n - 1} \dfrac{58}{\pi \sqrt{ {n}^{2} - {k}^{2} } }$

$\rm :\longmapsto\:y \: = \: \lim_{n \to \infty}\sum \limits_{k = 0}^{n - 1} \dfrac{58}{n\pi \sqrt{1- \dfrac{ {k}^{2} }{ {n}^{2} } } }$

Now, using Limit as a sum of definite integrals, we get

$\rm :\longmapsto\:y = \dfrac{58}{\pi} \displaystyle\int_0^1\rm \frac{dx}{ \sqrt{1 - {x}^{2} } }$

$\rm :\longmapsto\:y = \dfrac{58}{\pi} \bigg( {sin}^{ - 1}x \bigg) _0^1\rm$

$\rm :\longmapsto\:y = \dfrac{58}{\pi} \bigg( {sin}^{ - 1}1 - {sin}^{ - 1}0\bigg)$

$\rm :\longmapsto\:y = \dfrac{58}{\pi} \times \dfrac{\pi}{2}$

$\bf\implies \:y = 29$

Hence,

$\rm :\longmapsto\:\lim_{n \to \infty} \bigg( \sum \limits^{n + 9}_{k = 10}\frac{2^{\frac{11(k - 9)}{n} } }{ \log_{2}e^{ \frac{n}{11} } } - \sum \limits_{k = 0}^{n - 1} \dfrac{58}{\pi \sqrt{(n - k)(n + k)} } \bigg )$

$\rm \:  =  \: x - y$

$\rm \:  =  \: 2047 - 29$

$\rm \:  =  \: 2018$

So,

$\boxed{\tt{ \lim_{n \to \infty} \bigg( \sum \limits^{n + 9}_{k = 10}\frac{2^{11 \frac{k - 9}{n} } }{ \log_{2}e^{ \frac{n}{11} } } - \sum \limits_{k = 0}^{n - 1} \frac{58}{\pi \sqrt{(n - k)(n + k)} } \bigg ) = 2018}}$

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$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$