A) \[2({{2}^{n}}-1)+8n\]
B) \[2({{2}^{n}}-1)+6n\]
C) \[3({{2}^{n}}-1)+8n\]
D) \[4({{2}^{n}}-1)+8n\]
Correct Answer: D
Solution :
| [d] Let nth term of series is \[{{T}_{n}}\] then |
| \[{{S}_{n}}=12+16+24+40+....+{{T}_{n}}\] |
| Again \[{{S}_{n}}=12+16+24+....+{{T}_{n}}\] |
| On subtraction |
| \[0=(12+4+8+16+....+upto\,\,n\,\,terms)\,-{{T}_{n}}\] |
| or \[{{T}_{n}}=12+[4+8+16+...+\,upto\,\,(n-1)terms]\] |
| \[=12+\frac{4({{2}^{n+1}}-1)}{2-1}={{2}^{n-1}}+8\] |
| On putting n = 1, 2, 3?? |
| \[{{T}_{1}}={{2}^{2}}+8,\,{{T}_{2}}={{2}^{3}}+8,\,{{T}_{3}}={{2}^{4}}+8....\] etc. |
| \[{{S}_{n}}={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+....+{{T}_{n}}\] |
| \[=({{2}^{2}}+{{2}^{3}}+{{2}^{4}}+.....upto\,\,n\,\,terms)+(8+8+8+....\]\[\,\,upto\,\,n\,\,terms)\] |
| \[=\frac{{{2}^{2}}({{2}^{n}}-1)}{2-1}+8n=4({{2}^{n}}-1)+8n\]. |
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