A) \[\frac{1-(n+1){{x}^{n}}+n{{x}^{n+1}}}{{{(1-x)}^{2}}}\]
B) \[\frac{1-{{x}^{n}}}{1-x}\]
C) \[{{x}^{n+1}}\]
D) None of these
Correct Answer: A
Solution :
Let \[{{S}_{n}}\] be the sum of the given series to \[n\] terms, then \[{{S}_{n}}=1+2x+3{{x}^{2}}+4{{x}^{3}}+........+n{{x}^{n-1}}\] ?..(i) \[x{{S}_{n}}=\text{ }x+2{{x}^{2}}+3{{x}^{2}}+...........+n{{x}^{n}}\] ?..(ii) Subtracting (ii) from (i), we get \[(1-x){{S}_{n}}=1+x+{{x}^{2}}+{{x}^{3}}+.....\text{to}\] \[n\] terms \[-n{{x}^{n}}\] \[=\left( \frac{(1-{{x}^{n}})}{(1-x)} \right)-n{{x}^{n}}\] \[\Rightarrow {{S}_{n}}=\frac{(1-{{x}^{n}})-n{{x}^{n}}(1-x)}{{{(1-x)}^{2}}}=\frac{1-(n+1){{x}^{n}}+n{{x}^{n+1}}}{{{(1-x)}^{2}}}\].
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