A) \[{{2}^{n}}\]
B) \[n.\,\,{{2}^{n}}\]
C) \[n.\,\,{{2}^{n-1}}\]
D) \[n.\,\,{{2}^{n+1}}\]
Correct Answer: C
Solution :
Trick: Put \[n=1,\,\,2,\,\,3,....\] \[{{S}_{1}}=1,\,\,{{S}_{2}}=2+2=4\] Now by alternate (c), put \[n=1,\,\,2\] \[{{S}_{1}}={{1.2}^{0}}=1,{{S}_{2}}={{2.2}^{1}}=4\]
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