A) \[\frac{1}{1-a}\]
B) \[\frac{1}{1+a}\]
C) \[\frac{1}{1+{{a}^{2}}}\]
D) \[\frac{1}{{{(1-a)}^{2}}}\]
Correct Answer: D
Solution :
Given series, \[1+2a+3{{a}^{2}}+4{{a}^{3}}+....\] is a arithmetic geometric series. Here, \[{{a}_{1}}=1,\,\,d=1,\,\,r=a\] \[\therefore \] \[{{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}+\frac{d.r}{{{(1-r)}^{2}}}\] \[=\frac{1}{1-a}+\frac{1.a}{{{(1-a)}^{2}}}\] \[=\frac{1}{{{(1-a)}^{2}}}\]
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