Question

by the method of mathematical induction prove 10^n+3(4^n+2)+5 is divisible by 9

Answer 1

is that 4^(n+2)?

Answer 2

if not then it is not even true for n = 1

Answer 3

@meejoy ?

Answer 4

let n = 1
10+3(6) + 5 = 33
9 does not divide 33

Answer 5

Let n=1
\[LHS=10^1+3(4^{1+2})+5\]
\[=10+192+5\]
\[=207\]
which is divisible by 9.
Therefore true for n=1
Assume the formula is true for n=k
\[\large 10^k+3(4^{k+2})+5=9m\]
where
\[\large m \epsilon \mathbb{R}\]
For n=k+1
\[\large 10^{k+1}+3(4^{k+1+2})+5\]
\[\large =10^{k+1}+3(4^{k+3})+5\]

Answer 6

@meejoy ??

Answer 7

no @zzr0ck3r You got the equation wrong...

Answer 8

I see

Answer 9

yes?? @zzr0ck3r ..

Answer 10

@Azteck : are you done with your solution?

Answer 11

\[\large =(10^k\times 10)+3(4^k \times 4^3)+5\]
\[\large =(10^k \times 10)+3(64)(4^k)+5\]

Answer 12

Not yet.

Answer 13

ok..

Answer 14

Forget the last line I previously did.

Answer 15

\[\large =10(10^k+3(4^{k+2})+5)-45-18(4^{k+2})\]

Answer 16

\[\large =10(10^k+3(4^{k+2})+5)-9[5+2(4^{k+2})]\]
No matter what you put as k, for k+1 will be in the same form since you're subtracting a multiple of 9.

Answer 17

Therefore you solved it by mathematical induction. YOu can write your concluding lines. I'm a bit tired now.

Answer 18

=10(10k+3(4k+2)+5)−45−18(4k+2) where this came from? im a bit confuse

Answer 19

I purposely rigged the equation so I could get the same form to your original equation.

Answer 20

That is still equal to the previous line before that.

Answer 21

I wanted to get rid of 10^k+1

Answer 22

So I factorised it. But if you factorise 10 out just to get rid of 10^k+1, you would have to subtract some values for the middle and last terms to balance that out. Here's a basic example.

Answer 23

\[5^{n+1}-3x+4=5(5^n -3x+4)+12x-16\]

Answer 24

I wanted to get rid of the 1 in the n+1 so it can just be 1.

Answer 25

If I took the 5 out, I would also affect the middle and last terms. So I have to do some rigging.

Answer 26

so it can just be n**

Answer 27

that was noe of you:)

Answer 28

noe?

Answer 29

nice*

Answer 30

lol sorry

Answer 31

No worries. And thanks mate.

Answer 32

@Azteck : im sorry/... peace :)

Answer 33

@azteck: im sorry...

Answer 34

Okay, are you reading what I wrote for you?

Answer 35

im reading it..

Answer 36

okay im done.. and then??

Answer 37

and then you can turn the [5+2(4^k+2)] into a pronumeral like b. and then state "where b is an integer"

Answer 38

owww my g.. my head is aching >_<

Answer 39

That's what mathematical induction does to you if you don't focus.

Answer 40

Have you done that?

Answer 41

i think u'll find this way easier

Answer 42

:(

Answer 43

lol the substitution method. Our teacher was telling us to do it what I did. some other classes do it your way @Tushara

Answer 44

meejoy.... there are many different ways to do induction questions... some ppl find some ways easier than other ways

Answer 45

sorry about my typing. i'm typing with one hand. eating with my other hand.

Answer 46

@Azteck i was taught both ways in highscool, its good to understand both ways

Answer 47

@Tushara : thanks.. i appreciated it .. and also to you @Azteck ..

Answer 48

Yeah. My friend was teaching it to me that way. I found it pretty easy to solve, but sometimes, I make more mistakes using that method somehow, so I use the other one. I might revert to your method.

Answer 49

No worries.

Answer 50

but i still dont understand.. huhu T_T

Answer 51

Mathematical induction requires some "rigging" in order to get your required form.

Answer 52

oh ok... there are so many different methods for induction... the more logic and explaining u do, the better u get at induction, cos u can start to think in many different ways... substituion is by far the easiest method

Answer 53

You need to learn to RIG an expression so that you can get to where you want that expression to go to. That sentence sounds confusing, but it makes sense.

Answer 54

And @Tushara pretty much hit the target. You need to be logical as well.

Answer 55

that is why some questions say: do not use the substitution method

Answer 56

\[10^{k+1}+3(4^{k+1+2})+5\\=10(10^k)+3(4)(4^{k+2})+5\\=(9+1)10^k+3(3+1)(4^{k+2})+5\\=9(10^k)+(10^k)+3(3)(4^{k+2})+3(1)(4^{k+2})+5\\=9(10^k)+(10^k)+9(4^{k+2})+3(4^{k+2})+5\\=[9(10^k)+9(4^{k+2})]+[10^k+3(4^{k+2})+5]\\=9[10^k+4^{k+2}]+[9m]\\=9\left(10^k+4^{k+2}+m\right) \]