induction question (1 Viewer)

[HSC :(]

hey
can someone plz get bak to me on this question?
use the principal of mathematical induction to prove that
5^n + 2(11^n) is a multiple of three for all positive integers n.

thanx

 

[Huy]

This looks right to me, but see what you think:

Prove by mathematical induction that:
5^n +2(11^n) is divisible by 3, for all positive integer values of n.

Step 1: Test the statement for n = 1:
5^1 + 2(11^1) = 27, which is divisible by 3
.'. True for n = 1

Step 2: Assume the expression is divisible by 3 for n = k:
.'. 5^k + 2(11^k) / 3 = M, where M is some integer

ie 5^k + 2(11^k) = 3M
ie 5^k = [3M - 2(11^k)]

Step 3: Prove the expression is divisible by 3 for n = k + 1:
ie 5^(k+1) + 2(11^(k+1)) = 5^k . 5^1 + 2(11^k) . 2(11^1)
= [3M - 2(11^k)] . 5 + 2(11^k) . 2(11), using our assumption:

5^k = [3M - 2(11^k)]

= 15M - 5 . 2(11^k) + 2(11^k) . 2(11), on expanding
= 15M + 2(11^k)[5+22], on rearranging
= 15M + 27 . 2(11), on adding brackets and rearranging
= 3M + 9 . 2(11^k), factorising by 3
= 3(M + 3(2(11^k)), taking out the common factor of 3, where k is also another integer.
= 3(M + 6(11^k)), which is divisible by 3 because M is an integer, from above, and M x 6(11^k) is an integer.

Therefore, if the statement is true for n = k, it is also true for n = k + 1.

Step 4: Hence, if the statement is divisible by 3 for n = k, it is also divisible by 3 for n = k + 1. It is divisible by 3 for n = 1, so it is divisible by 3 for n = 2. If it is divisible by 3 for n = 2, it is also true for n = 3, and so on, for all positive integers n.

 

[Rahul]

assume: 5^n + 2 x 11^n =3I, I = 1, 2, 3....

consider n=1,
5 + 22 = 27 = 3I, where I=9
.'. true for n=1

assume n=k,

consider n = k,
5^k + 2 x 11^k = 3I ** , I = 1,2,3...

consider n = k+1
5^(k+1) + 2 x 11^(k+1)=3I
=5^k x 5 + 2 x 11 x 11^k
=5.5^k + 22.11^k
=5(5^k +2.11^k) + 12.11^k
=5.3I + 3(4.11^k).....from **
=3(5I) + 3(4.11^k)
=3(5I + 4.11^k)
.'. true for n = k+1

.'.if true for n = k, then n = k+1 is true. n=1 is true, hence n=2 is also true, and then n=3, and so on...hence by MI, the statement 5^n + 2(11^n) has multiples of 3 for n=1,2,3...

edit/ u beat me to it huy

 

[Huy]

=5.5^k + 22.11^k
=5(5^k +2.11^k) + 12.11^k

I like this step, I was trying to do something like this with 'extracting' the additional "on the side" component.

 

[Rahul]

yeh thanks

you just need to incorporate S(k) result in the S(k+1) statement. you can work backwards, by placing the S(k) result in the S(k+1) statement(after factorisation) and working out an answer that way. thats the way i do it anyway.

 

[Rand]

Hey Rahul, are you Indian?

 

[Huy]

Originally posted by Rand
Hey Rahul, are you Indian?

Yes.

(Since when was my name Rahul)

 

[Rahul]

Originally posted by Rand
Hey Rahul, are you Indian?

yes i am indeed, why may you ask?

Originally posted by Huy
(Since when was my name Rahul)

funnniiiiii

 

[Huy]

Originally posted by Rahul
yes i am indeed, why may you ask?

Rahul and Rand, sitting in a tree..

 

[Rand]

Piss the fuck off Huy

I have a friend called Rahul...and hes not Indian, so just curious....

:O

 

[Huy]

Originally posted by Rand
I have a friend called Rahul...and hes not Indian, so just curious....

LOL no, I was talking about "BOS Rahul" and you, jokingly.

 

[Rahul]

not indian?!!?

must have some connections with an indian background...

edit/ second time in this thred huy, that we have posted simultaneously