Question :
Let Vr = (Ur+1)^2 - (Ur)^2 = [ (n-2)((2n-3)]/ (n^2 + 1) , where n>= 1 and Ur>0
Show that (1/2)<= Vr <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!
(Ur+1)^2 - (Ur)^2 = 2Ur + 1
not sure what that has to do with n.
I guess I don't understand what Vr and Ur represent.
Ur+1 , Ur denote r+1th and rth term of a sequence respectively.
Vr = Ur+1 - Ur
Sigma Vr= [(n-2)(2n-3)]/(n^2 + 1)
Ur>0 and n>=1
Let me post the question again
To show that (1/2) <= Vr < 2, let's break down the problem into two parts: proving the lower bound (1/2) and proving the upper bound (2).
1. Proving the lower bound (1/2):
We can rewrite Vr as follows:
Vr = [(n - 2)(2n - 3)] / (n^2 + 1)
To find the lower bound, we need to prove that (1/2) <= Vr.
First, let's consider the numerator:
(n - 2)(2n - 3)
Expanding the parentheses, we get:
2n^2 - 7n + 6
Now, let's consider the denominator:
(n^2 + 1)
Since n^2 + 1 > 1 for all positive n, we can say that:
Vr = [(n - 2)(2n - 3)] / (n^2 + 1) < (n - 2)(2n - 3)
(Note: This inequality holds true for all n >= 1.)
To find the lower bound, we want to find the minimum value of (n - 2)(2n - 3). We can do this by finding the critical point where the derivative is zero.
Differentiating (n - 2)(2n - 3) with respect to n gives us:
d/dn [(n - 2)(2n - 3)] = 4n - 7
Setting the derivative equal to zero:
4n - 7 = 0
4n = 7
n = 7/4
Since n >= 1, the minimum occurs at n = 7/4.
Substituting n = 7/4 into (n - 2)(2n - 3):
(7/4 - 2)(2 * (7/4) - 3) = (3/4)(1/2) = 3/8
Therefore, we have proved that:
Vr < 3/8 < 1/2 for all n >= 1.
2. Proving the upper bound (2):
To prove the upper bound, we need to show that Vr < 2.
We already have Vr = [(n - 2)(2n - 3)] / (n^2 + 1).
Observe that:
[(n - 2)(2n - 3)] / (n^2 + 1) < 2
is equivalent to:
[(n - 2)(2n - 3)] < 2(n^2 + 1)
Expanding both sides of the inequality:
2n^2 - 7n + 6 < 2n^2 + 2
Simplifying the inequality:
-7n + 6 < 2
Rearranging the inequality:
-7n < -4
Dividing both sides by -7 (note that we flip the inequality since we are dividing by a negative number):
n > 4/7
Since n >= 1, we have:
n >= 1 > 4/7
Therefore, for all n >= 1, we can conclude that:
Vr < 2
To summarize, we have proved that (1/2) <= Vr < 2 for all n >= 1 without using mathematical induction.