Please help me with my homework! ( induction problem)

[valenzuela]

Show by induction that (3^(4n+2))+ (5^(2n+1))is a multiple of 14.
First person to do it gets a special mention on my location for a week /images/graemlins/laugh.gif

edit: time limit is 4am today to get a special mention on my location.

[aeest400]

Pick random integer, plug it in, divide result by 14? Or does he want a more rigorous proof?

[valenzuela]

how do you do that? /images/graemlins/frown.gif

[Duke]

Show for n=1:

3^6+5^3=854 = 14*61 == multiple of 14

Assume for n.

Show for n+1:

3^(4(n+1)+2) + 5^(2(n+1)+1) = 3^(4n+6) + 5^(2n+3)

= 3^6 * 3^4n + 5^3 * 5^2n

= 729(3^4n) + 125(5^2n)

= 729 * 81n + 125 * 25n

= 62174n

= 14*4441n

QED

[valenzuela]

thanks ! /images/graemlins/laugh.gif

[Silent A]

[ QUOTE ]
Show for n=1:

3^6+5^3=854 = 14*61 == multiple of 14

Assume for n.

Show for n+1:

3^(4(n+1)+2) + 5^(2(n+1)+1) = 3^(4n+6) + 5^(2n+3)

= 3^6 * 3^4n + 5^3 * 5^2n

= 729(3^4n) + 125(5^2n)

= 729 * 81n + 125 * 25n

[/ QUOTE ]

shouldn't this be:

= 729 * 81^n + 125 * 25^n

= (9^3)(9^2n) + (5^3)(5^2n)

it's not obvious to me how I should continue

[Duke]

[ QUOTE ]
[ QUOTE ]
Show for n=1:

3^6+5^3=854 = 14*61 == multiple of 14

Assume for n.

Show for n+1:

3^(4(n+1)+2) + 5^(2(n+1)+1) = 3^(4n+6) + 5^(2n+3)

= 3^6 * 3^4n + 5^3 * 5^2n

= 729(3^4n) + 125(5^2n)

= 729 * 81n + 125 * 25n

[/ QUOTE ]

shouldn't this be:

= 729 * 81^n + 125 * 25^n

= (9^3)(9^2n) + (5^3)(5^2n)

it's not obvious to me how I should continue

[/ QUOTE ]

I screwed up.

It's not 81n it's 81^n, so we're left with less help than I thought we had. There's more mathnastics to navigate through.

Sorry about that!

[Silent A]

Try this ...

1st number (n=1) = 729 + 125 = 854 = 14*61

let a_1 = 729 and b_1 = 125

(n+1)th number = 81a_n + 25b_n (see above)

where a_n and b_n are the two terms from the nth number

(n+1)th number = 56a_n + 25a_n + 25b_n
= 56a_n + 25(a_n+b_n)
= 4*14*a_n + 25(a_n+b_n)

a_n = an integer
a_n+b_n = multiple of 14

therfore (n+1)th number is a multiple of 14

[Silent A]

OK, lets do it properly in one go ...

1) for n=0 then (3^(4n+2))+ (5^(2n+1)) = 3^2 + 5^1 = 14

2) let a = 3^(4n+2)=9^(2n+1) and b = 5^(2n+1)

3) for n the equation becomes = a + b

4) for n+1 the equation becomes ...
= 3^(4(n+1)+2) + 5^(2(n+1)+1)
= 9^(2n+3) + 5^(2n+3)
= 81*9^(2n+1) + 25*5^(2n+1)
= 81a + 25b
= 56a + 25a + 25b
= 4*14a + 25(a+b)

therefore, if a is an integer and a+b = a multiple of 14, then the answer of the equation for (n+1) is also a multiple of 14

a is an integer when n>=0
a+b = 14 for n=0
therefore, by induction, the answer is a multiple of 14 for all n>=0