Induction (1 Viewer)

velox

Retired
Joined
Mar 19, 2004
Messages
5,521
Location
Where the citi never sleeps.
Gender
Male
HSC
N/A
Yes its fairly easy, but im having a mental block looking at the solution to this problem

Use mathematical induction to prove that for every positive integer n, (2^(n+2)) + (3^(3n) is divisible by 5.
 

withoutaface

Premium Member
Joined
Jul 14, 2004
Messages
15,098
Gender
Male
HSC
2004
Assume that

2^(k+2)+3^3k=5p(where p is an integer)

RTP
2^(k+3)+3^(3k+3)=2(5p-3^3k)+27*3^3k=10p+25*3^3k=5(2p+5*3^3k)

∴ divisible by 5
 

velox

Retired
Joined
Mar 19, 2004
Messages
5,521
Location
Where the citi never sleeps.
Gender
Male
HSC
N/A
how we get the 5p into the 2nd algebra manipulation? I dont understand the setting out.....fuck this is pissing me off, i had it yesterday, now it seems like i have never done it.
 
Last edited:

acmilan

I'll stab ya
Joined
May 24, 2004
Messages
3,989
Location
Jumanji
Gender
Male
HSC
N/A
S(n): (2^(n+2)) + (3^(3n))

for n = 1, S(1) = 8 + 27 = 35 which is divisible by 5 hence S(1) is true

Assume true for n = k
s(k): (2^(k+2)) + (3^(3k)) = 5M (where M is any integer(

To prove n = k + 1 is also true

s(k+1):
(2^(k + 2 + 1)) + (3^(3k + 3))
= (2.2^(k + 2)) + (3^(3k).3^3)
= (2.(5M - 3^(3k)) + (27.3^(3k)) [since 2^(k+2) = 5M - 3^(3k) from s(k)]
= 10M - 2.3^3k + 27.3^3k
= 10M + 25.3^(3k)
= 5(2M + 5.3^(3k))
= 5N (where N = (2M + 5.3^(3k)) which is an integer since M and 3^3k are integers)

hence true for s(k+1)

the 5M comes in because its was substituted from s(k) into s(k+1)
 

acmilan

I'll stab ya
Joined
May 24, 2004
Messages
3,989
Location
Jumanji
Gender
Male
HSC
N/A
wrx said:
Technically i dont, cos we dont start til next yr. But yeah it is easy. Read first post.

Next year? By the end of first term in year 12 my school had done the first two topics, tested us in a topic test and i dropped the course - all this in the first term
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top