13a grave sry ......
i hav edited it......
let a1,a2,a3 be a sequence of real no.s such dat an+1 = an + (an^2 + 1)^1/2 then LIM n-> ∞ (an/2n-1) is
1. 1
2. 0
3. -1
4. 2
-
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42 Answers
11SKY JI KOI DIFFERENCE NAHIN HAI
BAS a0=0 WAS MISSED BY DEEPANSHU
1nahi yaar....
in deepanshu's expression, a(n+1) is in the right hand side also...
but in prophet bhaiya's eqn, it is only in the left hand side.
11K
AB QUESTION KE BAARE MEIN KYA KHYAL HAI
LETS SOLVE IT TOGETHER AS I AM UNABLE O SOLVE IT BY MY OWN
1okie.. thanx for confirming deepansh..
but shud tell u..
i wasted 7-8 mins of my life [3] thinking of way to separate a[n-1] and a[n] [2]
neways its ok :)
1woh bajumein tune options likhe hai kya jara barabar likh diyo ...
1prophet bhaiya... the one u latexed and the question by deep.. are diff naa... those a[n] terms.... plz confirm...
13kya karoon is ques ne meri jaan leli
i m 2 desperate 2 noe the soln.....
341Time for a hint:Subash had almost hit upon it.
let a_k = \cot \theta
Express ak+1 too in terms of cot
341ok deepansh shouldnt be made to wait long.
When we make the substitution mentioned in the previous post, we get
a_{k+1} = a_k + \sqrt{1+a_k^2} = \cot \theta + \csc \theta = \frac{1+\cos \theta}{\sin \theta} \\ \\ = \cot \left(\frac{\theta}{2} \right)
Can you take it from here?
13i got ak+1 - ak = cosec(theta)
then...???
basically i m tryin to apply Vn - Vn-1 method
341No already have it in front of you.
If a_k = \cot \theta , a_{k+1} = \cot \left( \frac{\theta}{2} \right)
That means if a_{0} = \cot \theta_0 we have
a_{1} = \cot \left(\frac{\theta_0}{2} \right, a_2 = \cot \left(\frac{\theta_0}{4} \right),...., a_k = \cot \left(\frac{\theta_0}{2^k} \right),...
So we have \lim_{n \rightarrow \infty} \frac{a_n}{2^{n-1}} = \lim_{n \rightarrow \infty} \frac{1}{2^{n-1}} \cot \left(\frac{\theta_0}{2^n} \right) = \frac{2}{\theta_0}
Here \theta_0 = \frac{\pi}{2}. So the limit is \frac{4} {\pi}
3414/pi is correct dont worry. they would have missed some coefficient while copy-pasting
1yes 1 is wrong .. cuz i had calculated the limit in a very reliable way to get
1.273239544735163
didnt want to post only the answer to spoil the question as i was also thinking about the method ..
4 / pi should come to be almost that no. above (why almost ? well, bcoz my method was not as reliable as the mathematical one....)
341This is a very nice problem [ok I am partial to this coz i got very excited when the crux idea for the solution hit me!]. It is worth spending some time over it
Let me latex it for the members (also adding a condition missed by deepansh:
Given a_0 = 0 and the recursive relation a_n = a_{n-1} + \sqrt{1+a_{n-1}^2}, find \lim_{n \rightarrow \infty} \frac{a_n}{2^{n-1}}
341also if someone has seen this already [includes teachers and students] please let the students try it out for a while before jumping in with hints/solutions