Asked by Ashley
Question :
Let Vr = (Ur+1)^2 - (Ur)^2 , where Ur+1 & Ur denote the r+1th and rth term of a sequence where Ur>0
It is given that Sigma(from r=1 to n) Vr = [ (n-2)((2n-3)]/ (n^2 + 1) , where n>= 1 and Ur>0
Show that (1/2)<= (Ur+1)^2 - (Ur)^2 <2
My first though was to apply limits, resulting the right side of the equation and then appying Mathematical Induction gave the left hand side of the inequality.
Could you please guide me through a possible method of solving the above simulataneously, or without using principles of Mathematical Induction.
Or what could possibly be the most appropriate method for solving this?.
Thanks!
Let Vr = (Ur+1)^2 - (Ur)^2 , where Ur+1 & Ur denote the r+1th and rth term of a sequence where Ur>0
It is given that Sigma(from r=1 to n) Vr = [ (n-2)((2n-3)]/ (n^2 + 1) , where n>= 1 and Ur>0
Show that (1/2)<= (Ur+1)^2 - (Ur)^2 <2
My first though was to apply limits, resulting the right side of the equation and then appying Mathematical Induction gave the left hand side of the inequality.
Could you please guide me through a possible method of solving the above simulataneously, or without using principles of Mathematical Induction.
Or what could possibly be the most appropriate method for solving this?.
Thanks!
Answers
Answered by
oobleck
I don't like this
V1 = U2²-U1²
V2 = U3²-U2²
...
Vn = Un+1²-Un²
------------------------ add them up, and the sum from 1..n Vr telescopes to
Un+1²-U1²
Now we have
V1 = (1-2)(2-3)/(1+1) = 1/2
V1+V2 = (2-2)(4-3)/(4+1) = 0 = 1/2 + V2 ... V2 = -1/2
V1+V2+V3 = (3-2)(6-3)/(9+1) = 3/10 = -1/2 + V3 ... V3 = 4/5
Maybe you can use induction on this, but V2 fails the condition.
V1 = U2²-U1²
V2 = U3²-U2²
...
Vn = Un+1²-Un²
------------------------ add them up, and the sum from 1..n Vr telescopes to
Un+1²-U1²
Now we have
V1 = (1-2)(2-3)/(1+1) = 1/2
V1+V2 = (2-2)(4-3)/(4+1) = 0 = 1/2 + V2 ... V2 = -1/2
V1+V2+V3 = (3-2)(6-3)/(9+1) = 3/10 = -1/2 + V3 ... V3 = 4/5
Maybe you can use induction on this, but V2 fails the condition.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.