Question

the nth no of a sequence is a(n)= 5^n + 12n - 1.
Prove that it is divisible by 16 (using induction), considering a(n+1) - 5a(n).

Answer 1

so i proved a(1), got a(n+1) and 5a(n). what do i do now??

Answer 2

what is \(a_n\) ?

Answer 3

5^n + 12n - 1.

Answer 4

i just don't understand the last part of the q.

Answer 5

i am sorry, i meant what is \(a_{n+1}\) ?

Answer 6

btw, hi satellite :D

Answer 7

hi!
the idea is there will be some algebra involved to get \(a_{n+1}\) with part of it being \(a_n\) and the rest being 16 times something

Answer 8

ohh :P umm, so, i'm saying that a(n) = 16m.
so its 4(4m + 5^k + 3), with n = k, and substituting a couple of other things.

Answer 9

let me see if i can do the algebra

Answer 10

ohh, okay, go ahead :)

Answer 11

ok it is like this:

Answer 12

shoot.

Answer 13

it is really a bunch of algebra, that is all
you want to have a term that looks like \(5^n+12n-1\) when you compute
\[5^{n+1}+12(n+1)-1\]so you force it in there

Answer 14

here we go
\[5^{n+1}+12n+12-1=5\times 5^n+12n+11\]
\[=5(5^n+12n-1)+\text{stuff}\] so we just need to find out what the stuff is that makes it work
now
\[5(5^n+12n-1)=5^{n+1}+60n-5\] but you have
\[5^{n+1}+12n+11\] so you have to subtract off \(56n\) and add \(16\) to make it match
therefore you can say
\[5^{n+1}+12(n+1)-1=5(5^n+12n-1)-56n+16\]

Answer 15

my algebra is good, but my arithmetic is bad sorry
it should be "subtract off \(48n\) and add 16

Answer 16

so by this trick of algebra you get
\[5^{n+1}+12(n+1)-1=5(5^n+12n-1)-48n+16\]

Answer 17

therefore you have
\[a_{n+1}=5(a_n)-16(3n-1)\]
\(a_n\) is divisible by 16 "by induction" and \(16(3n-1)\) is divisible by 16 from your eyeballs, so the whole thing must be divisible by 16

Answer 18

yup, i'm working it out.

Answer 19

Thank youu SOO muchh. And it's good to see you (figuratively :P), cause i haven't been around in a while.