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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2006 · Mathematics · Sequences and Series

Let A_n= ((3)/(4))- ((3)/(4))^2+((3)/(4))^3+ +(-1)^ n-1 ((3)/(4))^n...

Let An=A_n=\left(\frac{3}{4}\right)-(34)2+(34)3+…+(−1)n−1(34)n\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n and Bn=1−AnB_n=1-A_n. Find a least odd natural number n0n_0, so that Bn>An,∀n≥n0B_n>A_n, \forall n \geq n_0.

Correct answer

7

Step-by-step solution

Bn=1−An>An⇒An<12B_n=1-A_n > A_n \Rightarrow A_n < \frac{1}{2} Now, An=34(1−(−34)n)1+34<12⇒(−34)>−16A_n=\frac{\frac{3}{4}\left(1-\left(-\frac{3}{4}\right)^n\right)}{1+\frac{3}{4}} < \frac{1}{2} \Rightarrow\left(-\frac{3}{4}\right) > -\frac{1}{6} Obviously, it is true for all even values of nn. But, for n=1,34<16n=3,2764<16n=5,2431024>16\begin{aligned} n & =1, \frac{3}{4}& < \frac{1}{6} \\ n & =3, & \frac{27}{64} & < \frac{1}{6} \\ n & =5, & \frac{243}{1024} & >\frac{1}{6} \end{aligned} which is true for n=7n=7. Obviously, n0=7n_0=7
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