Sequences And Series Ques 16
- For a positive integer $n$ let $a(n)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{\left(2^n\right)-1}$, then
$(1999,3 M)$
(a) $a(100) \leq 100$
(b) $a(100)>100$
(c) $a(200) \leq 100$
(d) $a(200)>100$
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Answer:
Correct Answer: 16.(a,d)
Solution: (a,d) Given, $a(n)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{2^n-1}$
$ \begin{aligned} & =1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\ldots+\frac{1}{7}\right)+\left(\frac{1}{8}+\ldots+\frac{1}{15}\right) \\ & +\ldots+\left(\frac{1}{2^{n-1}}+\ldots+\frac{1}{2^n-1}\right) \\ & <1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\ldots+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\ldots+\frac{1}{8}\right) \\ & +\ldots+\left(\frac{1}{2^{n-1}}+\ldots+\frac{1}{2^{n-1}}\right) \\ & =1+\frac{2}{2}+\frac{4}{4}+\frac{8}{8}+\ldots+\frac{2^{n-1}}{2^{n-1}} \\ & =1+1+1+1+\ldots+1=n \\ \end{aligned} $
Thus, $a(100) \leq 100$
Therefore, (a) is the answer.
$ \begin{aligned} & \text { Again, } a(n)=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\ldots+\frac{1}{8}\right) \\ & \quad+\ldots+\left(\frac{1}{2^{n-1}+1}+\ldots+\frac{1}{2^n}\right)-\frac{1}{2^n} \\ &>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\ldots+\frac{1}{8}\right) \\ &+\ldots+\left(\frac{1}{2^n}+\ldots+\frac{1}{2^n}\right)-\frac{1}{2^n} \\ &= 1+\frac{1}{2}+\frac{2}{4}+\frac{4}{8}+\ldots+\frac{2^{n-1}}{2^n}-\frac{1}{2^n} \\ &=1+\frac{1}{2}+\frac{1}{2}+\ldots+\frac{1}{2}-\frac{1}{2^n}=\left(1-\frac{1}{2^n}\right)+\frac{n}{2} \end{aligned} $
Therefore, $a(200)>\left(1-\frac{1}{2^{200}}\right)+\frac{200}{2}>100$
Therefore, (d) is also the answer.
BETA
