another induction q (1 Viewer)

[ezzy85]

Prove 7ⁿ + 13ⁿ + 19ⁿ is a multiple of 13, if n is odd.
Thanks

 

[spice girl]

Originally posted by ezzy85
Prove 7ⁿ + 13ⁿ + 19ⁿ is a multiple of 13, if n is odd.
Thanks

Prove case for n=1
7+13+19 = 39 which is multiple of 13 *tick*

Assume case true for n= k (k is odd)

then let 7^k + 13^k + 19^k = 13M (M is a positive integer)

Then for n=k+2 (the next odd integer)
7^k+2 + 13^k+2 + 19^k+2
= 49*7^k + 169*13^k + 361*19^k
= 49*13M + 120*13^k + 312*19^k
=13*(49M + 120*13^k-1 + 24*19^k )
= integer multiple of 13

Thus proven by maths induction

 

[ezzy85]

Re: Re: another induction q

Originally posted by spice girl

Prove case for n=1
7+13+19 = 39 which is multiple of 13 *tick*

Assume case true for n= k (k is odd)

then let 7^k + 13^k + 19^k = 13M (M is a positive integer)

Then for n=k+2 (the next odd integer)
7^k+2 + 13^k+2 + 19^k+2
= 49*7^k + 169*13^k + 361*19^k
= 49*13M + 120*13^k + 312*19^k
=13*(49M + 120*13^k-1 + 24*19^k )
= integer multiple of 13

Thus proven by maths induction

how did you get from
49*7^k + 169*13^k + 361*19^k

to
= 49*13M + 120*13^k + 312*19^k

 

[ezzy85]

dont worry about it, i see it now.
thanks alot for that

 

[bobo123]

would it be possible to do this question if n was all real numbers?

 

[spice girl]

If n can be any real number, the expression won't be an integer, let alone divisible by something...

 

[bobo123]

sorry sorry
all real integers

or just positive integers

 

[spice girl]

Yup, the induction above shows it's true for all positive integers. All integers are real anyway.