in Quantitative Aptitude edited by
264 views
0 votes
0 votes
Let $u_{n+1}=2u_{n}+1 \;(n=0,1,2,\dots)$ and $u_{n}=0$. Then $u_{10}$ nearest to ________
in Quantitative Aptitude edited by
13.4k points
264 views

1 Answer

0 votes
0 votes

Given that,

  • $u_{n+1} = 2 u_{n} + 1 (n=0,1,2,\dots)$
  • $u_{n} = 0$

Now, put the various values of $n$ and see the pattern.

  • $n=0 \Rightarrow u_{1} = 2u_{0} + 1 = 1= 2^{1}-1$
  • $n=1 \Rightarrow u_{2} = 2u_{1} + 1 = 3= 2^{2}-1$
  • $n=2 \Rightarrow u_{3} = 2u_{2} + 1 = 7= 2^{3}-1$
  • $n=3 \Rightarrow u_{4} = 2u_{3} + 1 = 15= 2^{4}-1$

Now, we can generalize the pattern.

$u_{n} = 2^{n} – 1$

$\Rightarrow u_{10} = 2^{10} – 1$

$\Rightarrow u_{10} = 1024 – 1$

$\Rightarrow \boxed{u_{10} = 1023}$

Correct Answer $:1023$ 

10.3k points

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true