MX1: Induction (1 Viewer)

[beve]

This is probably simple, but as I'm studying for a test I thought I'd better post anyway:

Prove by Mathematical Induction that 5^n + 2 x 11^n is a multiple of 3.

i. Prove true for n = 1

so 5^1 + 2 x 11^n = 77.

77/3 doesn't give a whole number however, so do you continue with the proof or is it just not a multiple of 3?

 

[lyounamu]

This is probably simple, but as I'm studying for a test I thought I'd better post anyway:

Prove by Mathematical Induction that 5^n + 2 x 11^n is a multiple of 3.

i. Prove true for n = 1

so 5^1 + 2 x 11^n = 77.

77/3 doesn't give a whole number however, so do you continue with the proof or is it just not a multiple of 3?

5^1 + 2 . 11^1 = 27 which is divisible by 3.

Let 5^n + 2.11^n = 3m where m is an integer.

Now, assume that it is true for n=k
i.e. 5^k + 2 . 11^k = 3m
Hence, prove that it is also true for n = k+1
5^(k+1) + 2 . 11^(k+1) = 3t where t is any integer
L.H.S. = 5. 5^k + 22 . 11^k = 5(5^k 2.11^k) + 12. 11^k
= 15m + 12 . 11^k
= 3(5m + 4 . 11^k)
= 3t since 5m + 4 + 11^k is an integer.
= R.H.S.

Therefore, proved by the principle of the mathematical induction.

 

[beve]

*slaps head*


thanks man

 

[lyounamu]

*slaps head*


thanks man

Did you get this from the CSSA 2001 Trial paper? I am doing it now and this question (exact question) came out. Funny thing is, I was about to do that question when I saw your post.

 

[beve]

Did you get this from the CSSA 2001 Trial paper? I am doing it now and this question (exact question) came out. Funny thing is, I was about to do that question when I saw your post.

Yup. =)