A) \[{{S}_{n}}<\frac{\pi }{3\sqrt{3}}\]
B) \[{{S}_{n}}>\frac{\pi }{3\sqrt{3}}\]
C) \[{{T}_{n}}<\frac{\pi }{3\sqrt{3}}\]
D) \[{{T}_{n}}\ge \frac{\pi }{3\sqrt{3}}\]
Correct Answer: A
Solution :
| we have \[{{S}_{n}}=\sum\limits_{k=1}^{n}{\frac{n}{{{n}^{2}}+kn+{{k}^{2}}}}\] |
| And \[{{T}_{n}}=\sum\limits_{k=0}^{n-1}{\frac{n}{{{n}^{2}}+kn+{{k}^{2}}};n=1,2,3....}\] |
| For \[n=1,\]we get |
| \[{{S}_{1}}=\frac{1}{1+1+1}=\frac{1}{3}=0.3\]and \[{{T}_{1}}=\frac{1}{1+0}=1\] |
| Also\[\frac{\pi }{3\sqrt{3}}=\frac{\pi \sqrt{3}}{9}=\frac{3.14\times \,1.73}{9}\]\[=0.34\times 1.73=0.58\] |
| \[\therefore {{S}_{1}}<\frac{\pi }{3\sqrt{3}}<{{T}_{1}},\] |
| \[\therefore {{S}_{n}}<\frac{\pi }{3\sqrt{3}}\,\operatorname{and}\,{{T}_{n}}>\frac{\pi }{3\sqrt{3}}\] |
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