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 :
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 \[{{T}_{n}}=12+[4+8+16+...+\text{upto }(n-1)\] terms] \[=12+\frac{4({{2}^{n-1}}-1)}{2-1}={{2}^{n+1}}+8\] On putting \[n=1,\,2,\,3......\] \[{{T}_{1}}={{2}^{2}}+8\], \[{{T}_{2}}={{2}^{3}}+8\], \[{{T}_{3}}={{2}^{4}}+8......etc.\] \[{{S}_{n}}={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+....+{{T}_{n}}\] \[=({{2}^{2}}+{{2}^{3}}+{{2}^{4}}+....\text{upto}\,\,\,n\,\,\,\text{terms)}\]\[+(8+8+8+......\]upto n terms) \[=\frac{{{2}^{2}}({{2}^{n}}-1)}{2-1}+8n=4({{2}^{n}}-1)+8n.\]
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