Mathematics

# Let$\displaystyle { S }_{ n }=\sum _{ k=1 }^{ n }{ \dfrac { n }{ { n }^{ 2 }+kn+{ k }^{ 2 } } }$ and $\displaystyle { T }_{ n }=\sum _{ k=0 }^{ n-1 }{ \dfrac { n }{ { n }^{ 2 }+kn+{ k }^{ 2 } } }$for $n=1,2,3,..$ Then,

${ T }_{ n }<\dfrac { \pi }{ 3\sqrt { 3 } }$

##### SOLUTION
$S_n=\sum_{k=1}^{n}\dfrac{n}{n^2+kn+k^2}$

$S_n=\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+\frac{k}{n}+\left ( \frac{k}{n} \right )^2}$

$S_n=\int_{0}^{1}\dfrac{1}{1+x+x^2}$

$S_n=\dfrac{\pi}{3\sqrt{3}}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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