For a positive integer n,let a(n) = 1 + frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We analyze the sum $a(n) = \sum_{k=1}^{2^n-1} \frac{1}{k}$.By grouping terms,we can observe that $a(n) = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{

Vedclass · Accepted Answer

Both $A$ and $C$

Vedclass · Answer

(D) We analyze the sum $a(n) = \sum_{k=1}^{2^n-1} \frac{1}{k}$.By grouping terms,we can observe that $a(n) = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \dots + \frac{1}{7}) + \dots + (\frac{1}{2^{n-1}} + \dots + \frac{1}{2^n-1})$.Each group of the form $\sum_{k=2^{m-1}}^{2^m-1} \frac{1}{k}$ is greater than $\frac{1}{2}$.Since there are $n$ such groups,$a(n) > \frac{n}{2}$.For $n=200$,$a(200) > \frac{200}{2} = 100$.Also,it is a known property that $a(n) \le n$ for these sums.Thus,$a(100) \le 100$ and $a(200) > 100$ are both true.

For a positive integer $n$,let $a(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^n - 1}$. Then:

Similar Questions

The sum of the first $n$ terms of a sequence is given by $S_n = 3n^2 + 4n + 15$. If $T_r$ is the $r^{th}$ term of the sequence,then $T_3 - T_1$ is equal to

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ is formed again. Find the numbers.

What is the sum of the first $40$ terms of the series $1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 + 9 + \dots$?

Find the $7^{\text{th}}$ term in the following sequence whose $n^{\text{th}}$ term is $a_{n} = \frac{n^{2}}{2^{n}}$.

Let the first term of a series be $T_1=6$ and its $r^{\text{th}}$ term $T_r=3T_{r-1}+6^r$ for $r=2, 3, \ldots, n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}(n^2-12n+39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,then $n$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module