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 :
| [d] Taking the sequence \[3,\,\,9,\,\,27,\,\,81,.....\] |
| Its nth 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 nth term \[={{3}^{n}}-1\] Hence, nth 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 \[({{S}_{n}})=\Sigma (1-{{3}^{-n}})\] |
| \[=n-({{3}^{-1}}+{{3}^{-2}}+....+{{3}^{-n}})\] |
| \[=n-\frac{{{3}^{-1}}\{1-{{({{3}^{-1}})}^{n}}\}}{1-{{3}^{-1}}}=n-\frac{1}{2}(1-{{3}^{-n}})\] |
| \[=n+\frac{1}{2}({{3}^{-n}}-1).\] |
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