induction help (1 Viewer)

[kr73114]

1) 9^(n+2) - 4^n is divisible by 5
2) 3^3n + 2^(n+2) is divisible by 5
prove true by induction for all integers n

 

[Drongoski]

1) 9^(n+2) - 4^n is divisible by 5
2) 3^3n + 2^(n+2) is divisible by 5
prove true by induction for all integers n

Q1)

True for n = 1 since 9^(1+2) - 4^1 = 725 = divisible by 5

ASsume true for n = k >= 1

i.e 9^(k+2) - 4^k = 5m for some INTEGER m

.: 9^([k+1]+2) - 4^[k+1] = 9 x (9^(k+2) - 4^k + 4^k) - 4 x 4^k

= 9 x 5m - 4^k x (9-4)

= 9 x 5m - 5 x 4^k = 5 x ( 9m - 4^k) = 5 x INTEGER (i.e. divisible by 5)

.: if true for n = k, true for n = k+1

.: by mathematical induction, true for all integer n >= 1

You should be able to do Q2 now.

 

[blackops23]

1) i'm gonna skip step 1, so...
Assume TRUE for n = k

Therefore 9^(k+2) - 4^k = 5M, M being any integer

Prove for n=k+1
Therefore:
9^(k+3) - 4^(k+1)
= 9*9(k+2) - 4*4^k
=9(5M + 4^k) - 4*4^k --> From the assumption, 5M + 4^k = 9^(k+2)
=45M + 9*4^k - 4*4^k
=45M + 5*4^k
=5(9M + 4^k)

THEREFORE TRUE FOR N=K+1, TRUE FOR EVERYTHING

remember for divisibilities, when you do the n=k+1 proof, you are trying to get it in a form where the divisible integer is factored outside of the expression (in this case the 5, whilst the inside expression is a WHOLE NUMBER, in this case (9M + 4^k) which is a whole number as M integer

2) Assume true for n=k
3^3k + 2^(k+2) = 5M

Prove for n=k+1

therefore:
3^(3k+3) + 2^(k+3)
=3^3 * 3^3k + 2*2^(k+2)
=27*3^3k + 2*(5M - 3^3k) --> From the assumption, 2^(k+2) = 5M - 3^3k
=27*3^3k + 10M - 2*3^3k
=25*3^3k + 10M
=5(5*3^3k + 2M), divisible by 5 as stuff inside brackets is a integer

Once again you can see that to prove that it is divisible by 5 for n=k+1, get the final expression into the 5*(wat eva the hell you want as long as its an integer).

 

[Drongoski]

Good job blackops23.