Show that:

a)

b)

c)

d)

e)

f)

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- December 20th 2008, 07:16 AMSeaDivisibility 12
Show that:

a)

b)

c)

d)

e)

f) - December 20th 2008, 07:27 AMrunning-gag
Hi

I did not read them all but most of them at least can be proved by induction - December 20th 2008, 07:46 AMSea
- December 20th 2008, 08:35 AMSoroban
Hello, Sea!

Here's one of them ... a non-inductive proof.

Quote:

Show that: .

We have: .

. . .

. . .

. . .

. . .

Therefore: .

- December 20th 2008, 08:57 AMSea
Binomial theorem...:(...similar...a lot of question...a lot of binomial theorem...:(...very difficult...

But,

Thanks a million...;):)... - December 20th 2008, 09:22 AMSoroban
Hello, Sea!

I tried problem (e) by Induction.

. . It took more work than I anticipated . . .

Quote:

Verify . . . . True!

Assume

Add . to both sides:

. .

. .

. . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

The inductive proof is compete.

- December 20th 2008, 11:16 AMMoo
- December 20th 2008, 11:32 AMMoo
- December 21st 2008, 05:16 AMSea

We must show

We can write f(n) in the form

where M(7) means 'is a multiple of 7'

The first term = 0 (mod 7) so now we are reduced to finding

This can be written

and this completes the proof. We have shown

i.e., the expression on the left is a multiple of 7

and

Thanks Anthony...;):)... - December 21st 2008, 08:32 AMMoo
Gaaaaah T.T so you know congruences !

Quote:

a)

So

Quote:

b)

Quote:

c)

Quote:

d)

(the red parts are changed into an equivalent modulo ..) - December 21st 2008, 09:07 AMSea
I know congruences...:)...

I understand

and...

Thanks a million...;):)... - December 21st 2008, 11:05 AMSea
- December 21st 2008, 11:57 PMSeaQuote:

f)

and

I think...

Quote:

Now if n=2 we get 25 + 2 x 3 + 1 = 32 = M(8)

and if n=3 we get 125 + 2 x 9 + 1 = 144 = M(8)

and so we must prove

It will be more convenient to use f(n+1) but this does not alter the proof

We can ignore the first term as it will be divisible by 8 and look at

if n is even .......(1)

and if n is odd .......(2)

For (1) we have and putting n = 2m to make it even

= M(8) + 8 = 0 (mod 8) so OK if n is even

For (2) we have n odd and write it 2m+1 so we have

so OK if n is odd

and so for n either even or odd we have shown

and

Thanks Anthony...;):)... - December 23rd 2008, 02:27 PMJaneBennet

Hence, if is odd, and if is even,