A) 19/7
B) 1/5
C) ¼
D) None of these
Correct Answer: B
Solution :
[b] \[S=1+4x+7{{x}^{2}}+10{{x}^{3}}+........\] \[x.S=x+4{{x}^{2}}+7{{x}^{3}}+.......\] Subtract \[S(1-x)=1+3x+3{{x}^{2}}+3{{x}^{3}}+......\] \[S(1-x)=1+3x\left( \frac{1}{1-x} \right),\] \[\because \,\,\,|x|<1\] \[S=\frac{1+2x}{{{(1-x)}^{2}}}\] Given: \[\frac{1+2x}{{{(1-x)}^{2}}}=\frac{35}{16}\] \[\Rightarrow \,\,\,16+32x=35+35{{x}^{2}}-70x\,\,\,\Rightarrow \,\,\,x=\frac{1}{5},\frac{19}{7}\] But \[|x|<1,\] \[\therefore \,\,x=1/5\]You need to login to perform this action.
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