Given a1=12(a0+Aa0), a2=12(a1+Aa1)andan+1=12(an+Aan) for n ≥ 2, wher

1.	Given a1=12(a0+Aa0), a2=12(a1+Aa1)andan+1=12(an+Aan) for n ≥ 2, where a > 0, A > 0. Prove that an−√Aan+√A=(a1−√Aa1+√A)2n−1
Answer» Given $a_{1} = \frac{1}{2} (a_{0} + \frac{A}{a_{0}})$ , $a_{2} = \frac{1}{2} (a_{1} + \frac{A}{a_{1}}) a n d a_{n + 1} = \frac{1}{2} (a_{n} + \frac{A}{a_{n}})$ for n $\geq$ 2, where a > 0, A > 0. Prove that $\frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2 n - 1}$

Given a1=12(a0+Aa0), a2=12(a1+Aa1)andan+1=12(an+Aan) for n ≥ 2, where a > 0, A > 0. Prove that an−√Aan+√A=(a1−√Aa1+√A)2n−1

Answer»

Given $a_{1} = \frac{1}{2} (a_{0} + \frac{A}{a_{0}})$ ,

$a_{2} = \frac{1}{2} (a_{1} + \frac{A}{a_{1}}) a n d a_{n + 1} = \frac{1}{2} (a_{n} + \frac{A}{a_{n}})$ for n $\geq$ 2, where a > 0, A > 0. Prove that $\frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2 n - 1}$

Discussion

No Comment Found

Related InterviewSolutions

The number of ways of choosing 2 cards from a pack of 52 where atleast one face card is choosen
Let P(3,3) be a point on the hyperbola, x2a2−y2b2=1. If the normal to it at P intersects the x−axis at (9,0) and e is its eccentricity, then the ordered pair (a2,e2) is equal to
P speaks truth in 75% cases while Q in 90%.in what % r they likely to contradict each other in stating the same fact .? Do u think B is true Do
If f(x) has the second order derivative at x = c such that f '(c) = 0 and f ''(c) > 0, then c is a point of _______________.
An A.P. consists of 43 terms, if the sum of five middle-most terms is 195, then the sum of the A.P. is
Find the angle between two vectors a and b with magnitudes √3 and 2 respectively, having a.b=√6
If y={x}[x], then 3∫0y dx is equal to,where {.} and [.] are fractional part function and greatest integer function respectively.
The graph of f(x)=−|log(x−3)|+3 will be
1∫0ex⋅x(1+x)2dx is equal to
What is a set and list all formulas of set.

Given a1=12(a0+Aa0), a2=12(a1+Aa1)andan+1=12(an+Aan) for n ≥ 2, where a &gt; 0, A &gt; 0. Prove that an−√Aan+√A=(a1−√Aa1+√A)2n−1

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment

Given a1=12(a0+Aa0), a2=12(a1+Aa1)andan+1=12(an+Aan) for n ≥ 2, where a > 0, A > 0. Prove that an−√Aan+√A=(a1−√Aa1+√A)2n−1