A) \[{{S}^{2}}\]
B) \[\frac{{{S}^{2}}}{2S+1}\]
C) \[\frac{2S}{{{S}^{2}}-1}\]
D) \[\frac{{{S}^{2}}}{2S-1}\]
Correct Answer: D
Solution :
Given, \[\sum\limits_{n=0}^{\infty }{{{r}^{n}}=S}\] \[\Rightarrow 1+r+{{r}^{2}}+......+\infty =S\] \[\Rightarrow \] \[\frac{1}{1-r}=S\Rightarrow r=\frac{S-1}{S}\] Now, \[1+{{r}^{2}}+{{r}^{4}}+......+\infty =\frac{1}{1-{{r}^{2}}}\] \[=\frac{1}{1-{{\left( \frac{S-1}{S} \right)}^{2}}}=\frac{{{S}^{2}}}{{{S}^{2}}-{{(S-1)}^{2}}}\] \[=\frac{{{S}^{2}}}{{{S}^{2}}-({{S}^{2}}+1-2S)}=\frac{{{S}^{2}}}{2S-1}\]
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