induction (1 Viewer)

[DeeBee]

hi,
i typed this question in the binomial theorem thread, but then i read "withouttaface"'s rules..and well the picture freaked the hell out of me..so im sorry i didnt follow the rules. ok back to the topic..induction (must not get off topic)
can anyone explain to me how to do this question?:
use mathematical induction to show that 3^n + 7^n is divisible by 10 for all odd positive integers n>=1
any ideas?

 

[Templar]

For n=1, 3^1+7^1=10

If true for n=k, for n=k+2 (just odd numbers)
3^(k+2)+7^(k+2)
=(3+7)(3^(k+1)-3^(k)*7+.....) => uses the factorisation of a^n+b^n=(a+b)(a^(n-1)-a^(n-2)*b+a^(n-3)*b^2....-a*b^(n-2)+b^(n-1)), for odd n)
=10(3^(k+1)-3^(k)*7.....)

Hence 10|3^(k+2)+7^(k+2)

 

[DeeBee]

ummm...
is that a^n + b^n a rule thing? cause i've never seen anything like that before.
and do the * mean times?

 

[Templar]

ummm...
is that a^n + b^n a rule thing? cause i've never seen anything like that before.
and do the * mean times?

It's a rule, and * means multiplication.

Archman's going to point out that I didn't really use induction, but that's the only way I can think of right now...

 

[DeeBee]

ahh,
yeh cause at schoolwe were learning like..saying that it equaled to 10M where M is any integer, then from there we move it all around then sub in the 10M somehow,
however, all the questiosn i've done theres only one number thats to the power of n. not two...
but thanks for the help.

 

[Templar]

I think you might be able to prove the formula of a^n+b^n=(a+b)(......) with induction, and then apply it to 3^n+7^n to show it is divisible by 10. Sorry for not being able to provide an exact answer.

 

[DeeBee]

ok i reckon i got it, but this stupid booklet has no answers..
do you guys think this is right?
i'll write the question again:
use mathematical induction to show that 3^n + 7^n is divisible by 10 for all odd positive integers n=>1

so i put n equals 1 and it works...cause i get 10.
then i assume s(k) is true, then since its odd im gonna prove s(k+2) is true.
first i say that 3^k + 7^k =10M (where M is any integer)
then: 3^(k+2) + 7^(k+2)= 10P (where P is any integer)
=3^k * 3^2 + 7^k * 7^2
=3^k * 9 + 49(10M-3^k) {from 3^K + 7^k=10M}
={(3^k * 9) + (49 * -3^k)} + 49 * 10M
=3^k{9+(-49)} + 49 * 10M
=3^k * (9-49) + 49 * 10M
=-40 * 3^k + 49 * 10M
=10(-4 * 3^K + 49M)
=10P
therefore, s(k+2) is true, as its divisible by 10.

yes?

 

[Templar]

Yes, that is correct.

Somehow I think you learnt more from that than if I gave you the correct answer.

 

[DeeBee]

hahahah, possibily,
but anyway..can you just explain to me that rule please? i've never seen it before, and it would be an advantage if i knew it...

 

[DeeBee]

ok i have another question, lol sorry.
you know in the question when they say n>=1, if its n>=4, what do you do?
do yuo first prove it with n=4 and then does it make any change to the assuming of k, and then the k+1? or will it now be k +5? wait a sec...i think i just answered my own question...
hahah, its k+1 ay, because if n is 4 then the next number is oviously 5. my bad.
ok heres another puzzling question:
the terms u of n of a sequence are such that u of 1=1 and u of n=sqrt(2+(u of n-1)) for n>=2. use mathematical induction to show that u of n<2 for all positive integers n>=1

 

[Templar]

ok heres another puzzling question:
the terms u of n of a sequence are such that u of 1=1 and u of n=sqrt(2+(u of n-1)) for n>=2. use mathematical induction to show that u of n<2 for all positive integers n>=1

Well you prove for n=1, then assuming true for n=k, for n=k+1:

u{k+1}=sqrt(2+u{k}) which is less than sqrt(2+2)=2

Hence u{n}<2

As for the 'rule', it isn't really any rule. Expand (a+b)(a^(n-1)-a^(n-2)*b+a^(n-3)*b^2....-a*b^(n-2)+b^(n-1)) and you'll get a^n+b^n. Don't worry too much about it, I don't think it'll ever appear in any HSC stuff. I don't really know what's in the syllabus, so somethings I use things that aren't taught.

 

[DeeBee]

OMG!
i understood that!! hoooray! thanks heaps templar! your an effin genius!!

and i think maybe i shoudlnt touch the a^n + b^n thingy, just for now, i dont wanna get too confused, im only just catching on now.

thanks heaps man!!