Discrete Math
posted by Francesca on .
Use mathematical induction to establish the following formula.
n
Σ i² / [(2i1)(2i+1)] = n(n+1) / 2(2n+1)
i=1
Thanks for any helpful replies :)

check if true for n=1
LS = 1/(1(3)) = 1/3
RS = 1(2)/(2(3) = 2/6 = 1/3 , checks!
Assume it to be true for n = k
that is ....
1/(1(3)) + 4/(3(5)) = .... k^2/((2k1)(2k+1)) = k(k+1)/(2(2k+1))
prove then it must be true for n = k+1
that is
1/(1(3)) + 4/(3(5)) + .. + (k+1)^2/((2k+1)(2k+3)) = (k+1)(k+2)/(2(2k+3))
LS = k(k+1)/2(2k+1)) + (k+1)^2/((2k+1)(2k+3))
=[k(k+1)(2k+3) + 2(k+1)^2]/[2(2k+1)(2k+3)]
= (k+1)[k)2k+3) + 2(k+1)]/[2(2k+1)(2k+3)]
= (k+1)[2k^2 + 5k + 2]/[2(2k+1)(2k+3)]
= (k+1)(2k+1)(k+2)/[2(2k+1)(2k+3)]
= (k+1)(k+2)/(22k+3)
= RS
QED! 
6th line should have been
1/(1(3)) + 4/(3(5)) + .... + k^2/((2k1)(2k+1)) = k(k+1)/(2(2k+1)) 
Ok thank you for your helpful response! I have a couple of questions though. . .
Is the 15th line suppose to be '(k+1)(k+2)/(22k+3)'?
Also, the 16th line = RS, which is what exactly? 
line 15th, clearly a typo, was hoping you would realize it was
= (k+1)(k+2)/(2(2k+3))
in 1/(1(3)) + 4/(3(5)) + .. + (k+1)^2/((2k+1)(2k+3)) = (k+1)(k+2)/(2(2k+3))
there is a left side (LS) and a right side (RS) of the equation.
I started with LS and proved that it equals the RS, thus showing that the equation is true. 
Yea that's what I thought. . .Hey if you don't mind helping me further I have been working on this problem for a while and I am a bit stuck. IDK where to go from here or if I am doing it correctly:
Use mathematical induction to prove the truth of each of the following assertions for all n ≥1.
5^2n – 2^5n is divisible by 7
If n = 1, then 5^2(1)  2^5(1) = 7, which is divisible by 7. For the inductive case, assume k ≥ 1, and the result is true for n = k; that is 7  (5^2k + 2^5k). Use the assumption to prove n = k + 1, in other words, 5^(2(k + 1))  2^(5(k + 1)) is divisible by 7. Now,
5^(2(k + 1))  2^(5(k + 1))
= 5^(2k + 2)  2(5k + 5)
= 5^(2k) · 5^2  2^(5k) · 2^5
= 25 · 5^(2k)  32 · 2^(5k)
= IDK what to do from here. . .
Any suggestions? Thank you again! 
Any suggestions?