A) \[{{2}^{n}}\]
B) \[{{2}^{n}}+1\]
C) \[{{2}^{n}}-1\]
D) \[{{2}^{n-1}}\]
Correct Answer: C
Solution :
[c] put n=1 in the given alternatives. All the expressions other than (3) give a value greater then 1. or It is a G.P.S\[{{S}_{n}}=\frac{a({{r}^{n}}-1)}{(r-1)}=\frac{1({{2}^{n}}-1)}{(2-1)}={{2}^{n}}-1.\]
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