Prove by math induction that {5^(n+1)}+{4.6^n} - 9 is always divisible by 20?

Question

Prove by math induction that {5^(n+1)}+{4.6^n} - 9 is always divisible by 20?

Answer 1

Presumably this holds only for intergers or you couldn't prove it via induction.

First check it is true for n =1

5^2 + 4.6^1 - 9 = 20.6 (so this isn't true so you obviously cannot prove it)

Answer 2

let p(n)=5^(n+1)+4*6^n - 9

for n=0:

5^(1)+4*6^0 - 9=20, so 20 divides p(0).

induction step n->n+1:

Since 20 divides p(n), we just check 20 divides p(n+1)-p(n) to prove that 20 divides p(n+1):

p(n+1)-p(n)=

5^(n+1)-5^n+4*6^n-4*6^(n-1)-9+9=

4*5^n+^(n-1)+4*(5*6^(n-1))=

20*5^(n-1)+20*6^(n-1)

obviously divideable by 20. q.e.d.

I second bkelkar, please revise your basics of induction.

Answer 3

If the statement is denoted by p(n) and

If {5^(n+1)}+{4.6^n} - 9 = 20k then

{5^(n+2)}+{4.6^(n+1)} - 9

=5{5^(n+1)}+6{4.6^n} - 9

=5{20 - (4.6^n) +9} +6{4.6^n} - 9

=100 - (4.6^n) +36

=100 - 4{6^n - 9}

Now a power of 6 ends with 6 so 6^n - 9 ends with 5 hence { } is a multiple of 5 and 4{ } is a multiple of 20

So 100 - 4{6^n - 9} is a multile of 20 and so

So is {5^(n+2)}+{4.6^(n+1)} - 9

Hence p(n+1)  is true when p(n) is true but p(1) is true hence the statement is true for every n.

I did it but please revise your basics of induction.

Answer 4

For n=1:

5^2+4.6^1-9=25+24-9=40=20*2. So this is true for n=1

We assume that for n=k, the statement is true so:

5^(k+1)+ (4.(6^k))-9=20*l

From that assumption, we will prove that it is also true for n=k+1

so:

5^(k+2)+(4.(6^(k+1))-9=(5*(5^(k+1)))+(...

5*(5^(k+1))+6*(4*(6^(k)))-45+36=

5*(5^(k+1))+5*(4*(6^k)))+(4*(6^k)))-5*...

5*(5^(k+1)+ (4.(6^k))-9)+4*(6^k)+36=

5*20l+4*(6^k+9)

We know that 20=4*5.So if we want to prove 4*(6^k+9) is divisible by20, we have to prove it to be divisible by 4 and 5 in the same time.

We already know that it is divisible by 4 because its form is 4 * a, So now we only have to prove that 6^k+9 is divisible by 5.

To be divisible by 5, its last digit have to be 5 or 0

we also know that 6^k always end with a 6 and 6+9=15. So we already know that 6^k+9 ends with5, so it is divisible by 5--> 4*(6^k+9) is divisible by 20. suppose that it is 20m

so for n=k+1, we have the equation can be expressed as 20l+20m which is divisible by 20.

So it is true for n=1, and when it is true for n=k, it is also true for n=k+1, This proves that this statement is always true for all natural numbers

Answer 5

it's not true. Assuming you mean **evenly** divisible by 20, as any number is divisible by 20

It's not true for any integer that I can see. 4.6^n is always going to be a non-integer.

{5^(n+1)}+{4.6^n} - 9

n = 0

5^1 + 4.6^0 -9

5 -9 = -4

n = 1

5^2 + 4.6^1 - 9

25 + 4.6 - 9 = 18.6

n = 2

5^3 + 4.6^^2 = 9

125 + 21.16 -9

.