I dont normally encounter much trouble doing Proof by Induction questions but got stuck today on the following question in a practise paper im doing:

Prove by induction that 5^n + 4n + 7 is divisible by 4 for all positive integral n

I have no idea how your meant to get this into the "usual" form that proof by inductions start with. Can anybody offer any help?

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# Proof By Induction

Started by Holmes, May 19 2005 02:15 PM

6 replies to this topic

### #1

Posted 19 May 2005 - 02:15 PM

### #2

Posted 19 May 2005 - 02:22 PM

Sorry mate i dont know!!

### #3

Posted 19 May 2005 - 02:30 PM

Divisibility inductions dont start is the usual form.

First of all you put 1 through. so you get 5^1+4(1)+7

that is equal to 5+7+4=16

16 is divisible by 4 and therefore true for n=1

Assume n=k to be true then you get

5^k+4k+7/4 (the / means divides by)

so assuming n=k is true form a statement for n=k+1

5^(k+1)+4(k+1)+7/4

5^k +5^1+4k+4+7/4

5^k+4k+16/4

I dont think the 3rd part ofthis is right but the first two bits are

First of all you put 1 through. so you get 5^1+4(1)+7

that is equal to 5+7+4=16

16 is divisible by 4 and therefore true for n=1

Assume n=k to be true then you get

5^k+4k+7/4 (the / means divides by)

so assuming n=k is true form a statement for n=k+1

5^(k+1)+4(k+1)+7/4

5^k +5^1+4k+4+7/4

5^k+4k+16/4

I dont think the 3rd part ofthis is right but the first two bits are

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### #4

Posted 19 May 2005 - 02:39 PM

when you do assume true for n=k you should write

5^k+4k+7 = 4m i.e. where m is an integer, 5^k+4k+7 is divisible by a multiple of 4

for the purposes of the n=k+1 section next, this can be rewritten as 5^k = 4m - 4k - 7

onto n=k+1

5^(k+1)+4(k+1)+7

5.5^k+4(k+1)+7

using the above from n=k

(4m - 4k - 7).5 + 4(k+1)+7

20m - 20k - 35 + 4k + 4 + 7

20m - 16k + 24

which is divisble by 4 i.e. 4(5m - 4k + 6) so true for n=k+1

since true for n=k+1 when n=k and also true when n=1, true for all positive integral n

5^k+4k+7 = 4m i.e. where m is an integer, 5^k+4k+7 is divisible by a multiple of 4

for the purposes of the n=k+1 section next, this can be rewritten as 5^k = 4m - 4k - 7

onto n=k+1

5^(k+1)+4(k+1)+7

5.5^k+4(k+1)+7

using the above from n=k

(4m - 4k - 7).5 + 4(k+1)+7

20m - 20k - 35 + 4k + 4 + 7

20m - 16k + 24

which is divisble by 4 i.e. 4(5m - 4k + 6) so true for n=k+1

since true for n=k+1 when n=k and also true when n=1, true for all positive integral n

### #5

Posted 19 May 2005 - 02:42 PM

Yeh so take the first step from my post and the then the steps from Allan's and you've got it! Must go and revise.

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### #6

Posted 19 May 2005 - 03:03 PM

Cheers, problem solved. Much appreciated

### #7

Posted 19 May 2005 - 03:04 PM

Much appreciated help from Allan, I'm rubbish at Maths

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