induction qu's (1 Viewer)

[currysauce]

1 3 unit and 1 4 unit quesitons... try and pic the difference

If n is a positive even number, use induction to prove that n^2+2n is divisible by 8

i get to the end and i have 8p + 4k +8 now the 4k doesn't sit right, the only thing i could do is say that this = 8q (where q = p + 1/2k + 1)

thoughts?

2. Prove by mathematical induction that if n is an integer n >=2 and a>0, then

1/ a (a+1) ^n = 1/a - 1/a+1 - 1/(a+1)2 - ... - 1 / (a+1)^n

 

[Estel]

1. since k is even and thus k/2 is an integer, the expression is divisible is by 8.

2. 1/a(a+1)^n -1/(a+1)^n+1 [by induction hypothesis]
= [a+1-a]/a(a+1)^n+1
= 1/a(a+1)^n+1

 

[Affinity]

you're throwing things away a bit too fast..

let v(k) = (2k)^2 + 2*(2k)
v(1) = 2^2 + 2*2 = 8 which is divisible by 8.

suppose the proposition is true for n = 2(k-1),
Then

v(k-1) = [2(k-1)]^2 + 2*2(k-1) is divisible by 8
and
v(k) = (2k)^2 + 2*(2k) = v(k-1) + 8k
(check this) which is obviously a multiple of 8. etc.


2.)
check for n = 1

now suppose it's true for n = k-1,
then

1/a - 1/a+1 - 1/(a+1)^2 - ... - 1 / (a+1)^(k-1) = 1/ a (a+1) ^(k-1)
1/a - 1/a+1 - 1/(a+1)^2 - ... - 1 / (a+1)^k = 1/ a (a+1) ^(k-1) - 1/(a+1)^k
= (a+1)/ a (a+1) ^k - = a/ a (a+1) ^k
= 1 / a (a+1) ^k

etc.

 

[currysauce]

thanks guys....

i don't understand the second one, so i'll ask my teacher 2moro