A) \[{{3}^{n}}+8n+1\]
B) \[\frac{1}{2}[{{3}^{n}}+8n-1]\]
C) \[\frac{1}{2}({{3}^{n}}+8n+1)\]
D) None of the above
Correct Answer: B
Solution :
The sequence of differences between successive terms is\[2,\,\,6,\,\,18,\,\,54,...\] Clearly, it is a\[GP\]. Let\[{{T}_{n}}\]be the\[{{n}^{th}}\]term of the given series and\[{{S}_{n}}\]be the sum of its n terms. Then, \[{{S}_{n}}=5+7+13+31+85+...+{{T}_{n-1}}+{{T}_{n}}\] ... (i) \[{{S}_{n}}=5+7+13+31+...+{{T}_{n-1}}+{{T}_{n}}\] ... (ii) Subtracting (ii) from (i), we get \[O=5+[2+6+18+54+..+({{T}_{n}}-{{T}_{n-1}})]-{{T}_{n}}\] \[\Rightarrow \] \[O=5+2\cdot \frac{{{3}^{n-1}}-1}{3-1}-{{T}_{n}}\] \[\Rightarrow \] \[{{T}_{n}}=5+({{3}^{n-1}}-1)=4+{{3}^{n-1}}\] \[\therefore \] \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{T}_{k}}=}\sum\limits_{k=1}^{n}{(4+{{3}^{K-1}})}\] \[=\sum\limits_{k=1}^{n}{4+\sum\limits_{k=1}^{n}{{{3}^{k-1}}}}\] \[=4n+(1+3+{{3}^{2}}+...+{{3}^{n-1}})\] \[=4n+1\times \left( \frac{{{3}^{n}}-1}{3-1} \right)\] \[=4n+\left( \frac{{{3}^{n}}-1}{2} \right)\] \[=\frac{1}{2}({{3}^{n}}+8n-1)\]
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