I tried putting over common denominator and get:
(n^3 + 3n^2 + 2n) / (2n) * (-n^2-2n+2)
2. b) Show by mathematical induction that:
(1 - 1/4) (1 - 1/9) (1 - 1/16) ... (1 - 1/n^2) = n+1 / 2n, for n >= 2
I did p(1) and it is true
I assumed p(k) is true and moved onto p(k+1) to see if it is true. I'm stuck here.
My work:
Since p(1) true the left side is replaced with the right, I get:
n+1 / 2n = (n+1/2n) (1-1/(n+1)^2)
A lil' kick in the right direction and I should be able to solve this.
I very much hope you mean you did p(2)! You are told that this is only true for n>= 2.
I assumed p(k) is true and moved onto p(k+1) to see if it is true. I'm stuck here.
My work:
Since p(1) true the left side is replaced with the right, I get:
n+1 / 2n = (n+1/2n) (1-1/(n+1)^2)
A lil' kick in the right direction and I should be able to solve this.
Yes I meant p(2), typo on my part. =)
But is the answer your gave with two terms on each side, but are not identical, the answer? I tried multiplying it out and am not able to factor it so that I get the two groups that are on the left side of the equation.
Hmmm... the way I was taught was do perform the series with p(2) true, then assume* p(k) is true and go right ahead to p(k+1) and prove that.
I am capable of applying math induction on a simpler type of series, except this particular one I cannot seem to simplify the right hand side and get the left hand side.
Holy makarel! This was a lot simpler than I had thought. Somehow my brain has been jammed and I was unable to compute. What can you expect studying 6+ hours of math a day?
Oh boy wish me luck on my exams and thank you enormously for your help and patience!