Mathematical induction to prove conjecture?

Question

sum(n on top r=1 at bottom)1/(4r^2-1)=n/(2n+1)

I have proved 4 1 n got 1/3 for both

now it wants me to prove for k

which i got n's as k's

its when k+1 that i just cant seem to get and answer at all and the examples i have got to help me with are all basics where the r's are straights powers ie ^2 not fractions as I have and I am really confussed

HELP!!!!!!!!!

Answer 1

P(n) is given proposition.
k is a value of n
P(k) is the proposition that:-
k
∑ 1 / (4 k² - 1) = k / (2k + 1)
r = 1

Have to show that P(1) and P(k + 1) are true.

Consider P(1)
LHS = 1 / (4 - 1) = 1 / 3
RHS = 1 / (2 + 1) = 1 / 3
Thus P(1) is true

Consider P(k + 1)
k + 1
∑1 / [ (4 (k + 1)² - 1 ] = (k + 1) / (2k + 3)
1
Have now to show that this is true , starting from P(k):-
LHS
k + 1
∑1 / [ 4 (k + 1)² - 1 ]
r = 1
RHS
k / (2k + 1) + 1 / [ 4 (k + 1)² - 1 ]
k / (2k + 1) + 1 / (4k² + 8k + 3)
k / (2k + 1) + 1 / (2k + 1)(2k + 3)
(2k² + 3k + 1) / (2k + 1) (2k + 3)
(2k + 1)(k + 1) / (2k + 1) (2k + 3)
(k + 1) / (2k + 3

Thus P(k + 1) is true
P(1) is also true
Thus
P(k) is true and P(n) is true.

Answer 2

While the inductive step is possible without doing any manipulation on the form of the sum, it's easier to see the inductive step if you realize this is a telescoping series.

Let us decompose 1/(4r²-1) using the method of partial fractions. We have:

1/(4r²-1) = A/(2r-1) + B/(2r+1)
1 = (2r+1)A + (2r-1)B

Substituting r=1/2 yields 2A=1, so A=1/2. Similarly, substituting r=-1/2 yields -2B=1, so B=-1/2. Thus we have:

1/(4r²-1) = 1/2 (1/(2r-1) - 1/(2r+1))

So consider adding the first few terms:

For n=1, we have 1/2 (1/(2-1) - 1/(2+1)) = 1/2 (1-1/3)
For n=2, we have 1/2 (1/(2-1) - 1/(2+1) + 1/(4-1) - 1/(4+1)) = 1/2 (1/(2-1) - 1/(4+1)) = 1/2 (1-1/5)
For n=3, we have 1/2 (1/(2-1) - 1/(2+1) + 1/(4-1) - 1/(4+1) + 1/(6-1) - 1/(6+1)) = 1/2 (1/(2-1) -1/(6+1)) = 1/2 (1-1/7)

You see what's happening here? Each time n increments, the first term of the new summand cancels out the last term of the previous one. Thus the only terms remaining after n summands are the first term of the first summand (which is 1/2 * 1/(2-1) = 1/2), and the last term of the last summand (which is 1/2 * (-1/(2n+1)) = -(1/2)/(2n+1). Thus the sum of the first n terms must be 1/2 - (1/2)/(2n+1) = 1/2 (1-1/(2n+1)) = 1/2 ((2n+1)/(2n+1) - 1/(2n+1)) = 1/2 (2n/(2n+1)) = n/(2n+1), as the formula you were given claims.

Okay, enough with the introduction, let's do some induction. We wish to show that [r=1, n]∑1/(4r²-1) = n/(2n+1). First we note that per the algebra we did above, [r=1, n]∑1/(4r²-1) = 1/2 [r=1, n]∑(1/(2r-1) - 1/(2r+1)) and n/(2n+1) = 1/2 (1 - 1/(2n+1)). Therefore it suffices to show that [r=1, n]∑(1/(2r-1) - 1/(2r+1)) = 1 - 1/(2n+1). First, we consider the case where n=1:

[r=1, 1]∑(1/(2r-1) - 1/(2r+1)) = 1 - 1/3 = 2/3
1 - 1/(2(1)+1) = 1-1/3 = 2/3 ✓

Next, we suppose that this holds when n=k, and show that it holds when n=k+1:

[r=1, k+1]∑(1/(2r-1) - 1/(2r+1))
= [r=1, k]∑(1/(2r-1) - 1/(2r+1)) + 1/(2(k+1)-1) - 1/(2(k+1)+1)

Invoking our inductive hypothesis that [r=1, k]∑(1/(2r-1) - 1/(2r+1)) = 1 - 1/(2k+1):

= 1 - 1/(2k+1) + 1/(2(k+1)-1) - 1/(2(k+1)+1)
= 1 - 1/(2k+1) + 1/(2k+1) - 1/(2(k+1)+1)
= 1 - 1/(2(k+1)+1) ✓

Which is precisely the formula when n=k+1, so if the formula holds for k then it must also hold for k+1. By induction, it holds for all k. Q.E.D.

Answer 3

Ok, so you've proved the conjecture holds for n=1.

Your next step is to first assume it holds for n=k, and then show that if it does, it also holds for n=k+1.

(Let SUM mean the sum from 1 to k)
So we assume SUM(1/4r²-1)=k/(2k+1)

Now, SUM(to k+1)=SUM + (1/(4(k+1)²-1)) (you see i've just added the k+1 term separately)

This gives SUM(to k+1)=k/(2k+1)+1/(4k²+8k+3) (from our assumption and expanded of the square in the k+1 term)

Therefore, by combining the fractions, we have:
SUM(to k+1)=(4k^3+8k²+5k+1)/(8k^3+20k²+14k+3)

Now you have to find a common factor to cancel this down. To do this, I assumed k+1 was a factor of the top (as we should expect). This gave a common factor of 4k²+4k+1, and our fraction cancelled down to (k+1)/(2k+3)=(k+1)/(2(k+1)+1)
thus proving the conjecture.