A) \[n-\frac{1}{2}({{3}^{n}}-1)\]
B) \[n+\frac{1}{2}({{3}^{n}}-1)\]
C) \[n-\frac{1}{2}(1-{{3}^{-n}})\]
D) \[n+\frac{1}{2}({{3}^{-n}}-1)\]
Correct Answer: D
Solution :
| Taking the sequence 3, 9, 27, 81...... |
| Its \[{{n}^{th}}\] term \[=3{{(3)}^{n-1}}={{3}^{n}}\] |
| Also take the sequence 2,8,26, 80..... or \[(3-1),(9-1),(27-1),(81-1),....\] |
| Its \[{{n}^{th}}\] term \[={{3}^{n}}-1\] |
| Hence, \[{{n}^{th}}\] term of the sequence \[\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+.....\] is \[\frac{{{3}^{n}}-1}{{{3}^{n}}}\] or \[1-{{3}^{-n}}\] |
| Now the sum \[({{S}_{n}})=\Sigma (1-{{3}^{-n}})\] |
| \[=n-({{3}^{-1}}+{{3}^{-2}}+....+{{3}^{-n}})\] |
| \[=n-\frac{{{3}^{-1}}\{1-{{({{3}^{-1}})}^{n}}\}}{1-{{3}^{-1}}}=n-\frac{1}{2}(1-{{3}^{-n}})\] |
| \[=n+\frac{1}{2}\,({{3}^{-n}}-1).\] |
You need to login to perform this action.
You will be redirected in
3 sec
