edited by
1,384 views
3 votes
3 votes

Let $(v_n)$ be a sequence defined by $v_1 = 1$ and $v_{n+1} = \sqrt{v_n^2 +\left(\dfrac{1}{5}\right)^n}$ for $n\geq1$. Then $\displaystyle{\lim_{n \rightarrow \infty}v_n}$ is

  1. $\sqrt{5/3}$
  2. $\sqrt{5/4}$
  3. $1$
  4. $\text{non existent}$
edited by

2 Answers

Best answer
8 votes
8 votes
$V_{n+1} = \sqrt{V_{n}^2 + (1/5)^n}$

on  squaring both side, we get

$V_{n+1}^2 - V_{n}^2 = (1/5)^n$

$V_{2}^2 - V_{1}^2 = (1/5)$

$V_{3}^2 - V_{2}^2 =  (1/5)^2$

$V_{4}^2 - V_{3}^2 =  (1/5)^3$

$V_{5}^2 - V_{4}^2 =  (1/5)^4$

$\vdots$

$V_{n}^2 - V_{n-1}^2 = (1/5)^n-1$

On adding all we get,

$V_{n}^2 - V_{1}^2 = ((1/5) + (1/5)^2 + (1/5)^3 + (1/5)^4 + (1/5)^5 +\dots + (1/5)^{n-1}$

$\quad = ( 1 - (1/5)^{n-1}) / 4$

$\quad =  1/4 - (1/5)^{n-1}/4 $

$\implies V_{n}^2 =  V_{1}^2 +  1/4 - (1/5)^{n-1}/4 $

$\quad \quad  =  1+ 1/4 - (1/5)^{n-1}/4 $

$\quad \quad  = 5/4 - (1/5)^{n-1}/4 $

on taking limit, we get

$V_{n} = \sqrt{5/4}$

$$(v)\lim_{x\to \infty}a^x = \begin{cases}0, &0\leq a < 1 \\1, \quad &a= 1\\\infty, &a>1\\\text{does not exist}, & a < 0\end{cases}$$

Correct Answer: $B$
edited by
1 votes
1 votes
Vn=root(1+(1/5)+(1/5)^2+.......(1/5)^n-1)

lim n->infinite Vn=root((5/4)-(1/(4*5^(n-1))))  =root(5/4)

answer (B)
Answer:

Related questions

0 votes
0 votes
0 answers
1
go_editor asked Sep 15, 2018
365 views
Let $(v_n)$ be a sequence defined by $v_1=1$ and $v_{n+1}=\sqrt{v_n^2 +(\frac{1}{5})^n}$ for $n\geq1$. Then $\lim _{n\rightarrow \infty} v_n$ is$\sqrt{5/3}$$\sqrt{5/4}$$1...
0 votes
0 votes
0 answers
2
go_editor asked Sep 15, 2018
443 views
Let $(x_n)$ be a sequence of real numbers such that the subsequences $(x_{2n})$ and $(x_{3n})$ converge to limits $K$ and $L$ respectively. Then$(x_n)$ always convergesif...
0 votes
0 votes
2 answers
3
go_editor asked Sep 15, 2018
1,145 views
Let $S\subseteq \mathbb{R}$. Consider the statement “There exists a continuous function $f:S\rightarrow S$ such that $f(x) \neq x$ for all $x \in S.$ ”This statement ...
0 votes
0 votes
0 answers
4
Tesla! asked Apr 24, 2018
1,171 views
For $a,b \in \mathbb{R}$ and $b a$ , the maximum possible value of the integral $\int_{a}^{b}(7x-x^{2}-10)dx$ is$\frac{7}{2}\\$$\frac{9}{2}\\$$\frac{11}{2}\\$none of the...