If I could get some help on this as soon as possible, that would be great. :p

Here are the quesitons (I took a screenshot because I'm lazy like that :p)

http://img297.imageshack.us/img297/8387/mathho6.png

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- November 10th 2006, 07:17 PMAJLA Bit Of Help Please. Induction Arguments
If I could get some help on this as soon as possible, that would be great. :p

Here are the quesitons (I took a screenshot because I'm lazy like that :p)

http://img297.imageshack.us/img297/8387/mathho6.png - November 10th 2006, 07:57 PMSoroban
Hello, AJL!

Here's the first one . . .

Quote:

Verify . . . true!

Assume

Add to both sides:

. .

The right side is: .

Hence, we have: .

And we have proved . . . Therefore, is true.

- November 11th 2006, 04:24 AMtopsquark
2) is divisible by 3 for n = 1, 2, 3,...

The n = 1 case is trivial: .

So assume the problem is true for n = k. Then we need to show that

is divisible by 3.

Let for some integer m. Then we know that .

Now,

So .

Thus is divisible by 3.

If you know modular Mathematics:

(mod 3) then (mod 3). Thus (mod 3). Thus (mod 3).

Thus is also divisible by 3.

-Dan - November 11th 2006, 04:37 AMtopsquark
3)

For n = 1:

Check!

Assume this is true for some n = k. We need to show that

Well, the first k terms of this sum to according to assumption, so we need to show

As advertised.

-Dan - November 11th 2006, 04:40 AMtopsquark
HINT: Use the same technique on 4 that I used in 3.

HINT: The same comment applies for 5, but you need to add a lot of fractions to get there!

-Dan - November 11th 2006, 04:49 AMtopsquark
6)

For n = 3:

Check!

Assume this is true for some n = k. We need to show that:

<-- Divide both sides by 2

But we know that (You can verify this yourself.)

Thus

Thus (After multiplying both sides by 2, just to make things clearer.)

-Dan - November 11th 2006, 04:59 AMtopsquark
7) T(0) = 1 and T(n) = 2nT(n-1). Show that .

First, to get rid of the n = 0 case:

Check!

Now,

and

Check!

So assume this is true for n = k.

We need to show that

If the theorem is true for n = k, then . So we need to show that

Well, rearranging the LHS a bit:

as advertised.

-Dan - November 11th 2006, 05:18 AMSoroban
Hello again, AJL!

Quote:

,

Verify . . . True!

Assume

Multiply both sides by

. .

. .

Hence: .

Therefore, is true for all

- November 11th 2006, 09:51 AMearbothnr. 4 only