The Student Room Group

FP1 - [Proof]

U1=−1,U2=0,Un+2=6Un+1−9Un,ProveUn=(n−2)3n−1 U_1 = -1 , U_2 = 0 , U_n+2 = 6U_n+1 - 9U_n , Prove U_n = (n-2)3^{n-1}
So I done the first bit

Step 1
Prove true for n=1 and n = 2
U1=(1−2)30=1[br]U2=(2−2)31=0 U_1 = (1-2)3^0 = 1[br]U_2 = (2-2)3^1 = 0
Both true

Step 2
Assume true for n=k and n=k+1, if true for n=k and n=k+1 then true for n=k+2

Uk=(k−2)3k−1[br]Uk+1=(k−1)3k[br]ProveUk+2=(k)3k+1[br]Uk+2=6(k−1)3k−9(k−2)3k−1[br] U_k = (k-2)3^{k-1}[br]U_k+1 = (k-1)3^k[br]Prove U_k+2 = (k)3^{k+1}[br]U_k+2 = 6(k-1)3k-9(k-2)3^{k-1}[br]
.....
....
....
Step 1 + Step 2 proves true for all n by induction

I don't know what to do now, may someone help me please
(edited 7 years ago)

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Reply 1
Avatar for B_9710
B_9710
18
You have to show that Uk+2=k(3k+1) U_{k+2} =k(3^{k+1}) .
You are correct so far and you can get to the required result from the step you're at.
(edited 7 years ago)
Reply 2
Avatar for AdeptDz
AdeptDz
OP
21
Original post by B_9710
You have to show that Uk+2=k(3k+1) U_{k+2} =k(3^{k+1}) .


Yeh i know that bit thanks, i just don't know how to make [br]Uk+2=6(k−1)3k−9(k−2)3k−1 [br]U_k+2 = 6(k-1)3k-9(k-2)3^{k-1}
show
Uk+2=k(3k+1) U_{k+2} =k(3^{k+1})
Reply 3
Avatar for AdeptDz
AdeptDz
OP
21
Original post by B_9710
You have to show that Uk+2=k(3k+1) U_{k+2} =k(3^{k+1}) .
You are correct so far and you can get to the required result from the step you're at.


I expanded it but i remember my teacher saying don't expand it unless you absolutely have to, and i can't tell if i have to or not like if there is another option
You know you need a 3k+13^{k+1}, so look for ways to get that. What can you do with 6 and 9?
Reply 5
Avatar for AdeptDz
AdeptDz
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21
Original post by NotNotBatman
You know you need a 3k+13^{k+1}, so look for ways to get that. What can you do with 6 and 9?


take out a factor of 3?
(edited 7 years ago)
Original post by AdeptDz
divide them by 2 and 3? to get 3?


Don't divide, but factorise. Remember you can't change the value.

9 can be written as 3^x right?

And then what happens with the 3?
Reply 7
Avatar for AdeptDz
AdeptDz
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Original post by NotNotBatman
Don't divide, but factorise. Remember you can't change the value.

9 can be written as 3^x right?

And then what happens with the 3?


Yeh my bad, i editted the post after you checked it im guessing

ohh, let me try see what i get now
Thanks
Reply 8
Avatar for AdeptDz
AdeptDz
OP
21
So would it be
Uk+2=6(k−1)3k−9(k−2)3k−1[br]therefore[br]=[br]6(k−1)3k−32(k−2)3k−1[br][br]therefore[br]=3[2(k−1)k−3(k−2)3k−2][br][br] U_k+2 = 6(k-1)3k-9(k-2)3^{k-1} [br]therefore[br]= [br]6(k-1)3k-3^2(k-2)3^{k-1} [br][br]therefore [br]= 3[2(k-1)k-3(k-2)3^{k-2}][br][br]
Original post by AdeptDz
So would it be
Uk+2=6(k−1)3k−9(k−2)3k−1[br]therefore[br]=[br]6(k−1)3k−32(k−2)3k−1[br][br]therefore[br]=3[2(k−1)k−3(k−2)3k−2][br][br] U_k+2 = 6(k-1)3k-9(k-2)3^{k-1} [br]therefore[br]= [br]6(k-1)3k-3^2(k-2)3^{k-1} [br][br]therefore [br]= 3[2(k-1)k-3(k-2)3^{k-2}][br][br]


Rather than taking factorising 3 in the last step, do you notice that you'll get
in the first term 3×3k3 \times 3^k and in the second term there is a 32×3k3^2 \times 3^k remembering that ab×ac=ab+ca^b \times a^c =a^{b+c}

Also you've written it a bit wrong. It is
uk+2=6(k−1)×3k−9(k−2)×3k−1 u_{k+2} = 6(k-1)\times 3^{k} - 9(k-2) \times 3^{k-1}

remember it's 3^k, not 3k, so you can't factorise 3 like that.
(edited 7 years ago)
Reply 10
Avatar for AdeptDz
AdeptDz
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21
Original post by NotNotBatman
Rather than taking factorising 3 in the last step, do you notice that you'll get
in the first term 3×3k3 \times 3^k and in the second term there is a 32×3k3^2 \times 3^k remembering that ab×ac=ab+ca^b \times a^c =a^{b+c}

Also you've written it a bit wrong. It is
uk+2=6(k−1)×3k−9(k−2)×3k−1 u_{k+2} = 6(k-1)\times 3^{k} - 9(k-2) \times 3^{k-1}

remember it's 3^k, not 3k, so you can't factorise 3 like that.


cool so:
6(3k+1−3k)−3k+2+18∗3k−1 6(3^{k+1}-3^k) - 3^{k+2} + 18 * 3^{k-1}
Original post by AdeptDz
cool so:
6(3k+1−3k)−3k+2+18∗3k−1 6(3^{k+1}-3^k) - 3^{k+2} + 18 * 3^{k-1}


The adding power rule only applies when it's the same base, so k×3k≠3k+1 k \times 3^k \neq 3^{k+1}

3k×31=3k+1 3^k \times 3^1 = 3^{k+1}

So you want on the first term; 2[31(k−1)×3k]=2(k−1)×31×3k2[3^1(k-1)\times 3^{k}] = 2(k-1) \times 3^1 \times 3^k, so the powers of 3 can be added. do the same with 9=329=3^2

But leave the brackets (k-1) and (k-2) for now.
(edited 7 years ago)
Reply 12
Avatar for AdeptDz
AdeptDz
OP
21
Original post by NotNotBatman
The adding power rule only applies when it's the same base, so k×3k≠3k+1 k \times 3^k \neq 3^{k+1}

3k×31=3k+1 3^k \times 3^1 = 3^{k+1}

So you want on the first term; 2[31(k−1)×3k]=2(k−1)×31×3k2[3^1(k-1)\times 3^{k}] = 2(k-1) \times 3^1 \times 3^k, so the powers of 3 can be added. do the same with 9=329=3^2

But leave the brackets (k-1) and (k-2) for now.


oh my bad
Would it be
..−32(k−2)3k−1 ..-3^2(k-2)3^{k-1}

so [br]2[31(k−1)×3k]−32(k−2)3k−1[br] 2[3^1(k-1)\times 3^{k}] - 3^2(k-2)3^{k-1}
(edited 7 years ago)
Original post by AdeptDz
oh my bad
Would it be
..−32(k−2)3k−1] ..-3^2(k-2)3^{k-1}]

so [br]2[31(k−1)×3k]−32(k−2)3k−1[br] 2[3^1(k-1)\times 3^{k}] - 3^2(k-2)3^{k-1}


yes, then you can add the powers of 3 on each individual term.
Reply 14
Avatar for AdeptDz
AdeptDz
OP
21
Original post by AdeptDz
oh my bad
Would it be
..−32(k−2)3k−1 ..-3^2(k-2)3^{k-1}

so [br]2[31(k−1)×3k]−32(k−2)3k−1[br] 2[3^1(k-1)\times 3^{k}] - 3^2(k-2)3^{k-1}


so
[br]2[3k+1(k−1)]−3k+1(k−2)[br][br]2[3^{k+1}(k-1)] - 3^{k+1}(k-2)[br]
Original post by AdeptDz
so
[br]2[3k+1(k−1)]−3k+1(k−2)[br][br]2[3^{k+1}(k-1)] - 3^{k+1}(k-2)[br]


Yes, then you can factorise.
Reply 16
Avatar for AdeptDz
AdeptDz
OP
21
Original post by NotNotBatman
Yes, then you can factorise.


(3k+1)(k−1)−(k+2) (3^{k+1})(k-1)-(k+2)
Got a feeling i did that wrong as idk where to put the 2
Original post by AdeptDz
(3k+1)(k−1)−(k+2) (3^{k+1})(k-1)-(k+2)
Got a feeling i did that wrong as idk where to put the 2


Factorise the whole thing, so use a big bracket and leave the 2 on the first term, it doesn't matter which way around it's multiplied.

3k+1[2(k−1)−(k+2)]3^{k+1}[2(k-1)-(k+2)] when you multiply out it's the same right?
Reply 18
Avatar for AdeptDz
AdeptDz
OP
21
Original post by AdeptDz
so
[br]2[3k+1(k−1)]−3k+1(k−2)[br][br]2[3^{k+1}(k-1)] - 3^{k+1}(k-2)[br]


Original post by NotNotBatman
Yes, then you can factorise.

1sec, let me have another go
Reply 19
Avatar for AdeptDz
AdeptDz
OP
21
[br]2[3k+1(k−1)]−3k+1(k−2)[br]2×3k+1(k−1)−3k+1(k−20)[br]3k+1(2(k−1)−(k−2))[br][br][br][br]2[3^{k+1}(k-1)] - 3^{k+1}(k-2)[br]2\times3^{k+1}(k-1) - 3^{k+1}(k-20)[br]3^{k+1}(2(k-1)-(k-2))[br][br][br]

there
(edited 7 years ago)

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