Join TSR now and get all your revision questions answeredSign up now

FP1 - [Proof] Watch

    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
     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
     U_1 = (1-2)3^0 = 1

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

     U_k = (k-2)3^{k-1}

U_k+1 = (k-1)3^k

Prove U_k+2 = (k)3^{k+1}

U_k+2 = 6(k-1)3k-9(k-2)3^{k-1}
    .....
    ....
    ....
    Step 1 + Step 2 proves true for all n by induction

    I don't know what to do now, may someone help me please
    Online

    3
    ReputationRep:
    You have to show that  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.
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by B_9710)
    You have to show that  U_{k+2} =k(3^{k+1}) .
    Yeh i know that bit thanks, i just don't know how to make  

U_k+2 = 6(k-1)3k-9(k-2)3^{k-1}
    show
     U_{k+2} =k(3^{k+1})
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by B_9710)
    You have to show that  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
    • Community Assistant
    Offline

    3
    ReputationRep:
    You know you need a 3^{k+1}, so look for ways to get that. What can you do with 6 and 9?
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by NotNotBatman)
    You know you need a 3^{k+1}, so look for ways to get that. What can you do with 6 and 9?
    take out a factor of 3?
    • Community Assistant
    Offline

    3
    ReputationRep:
    (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?
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    So would it be
      U_k+2 = 6(k-1)3k-9(k-2)3^{k-1} 

therefore

= 

6(k-1)3k-3^2(k-2)3^{k-1}   



therefore 

= 3[2(k-1)k-3(k-2)3^{k-2}]
    • Community Assistant
    Offline

    3
    ReputationRep:
    (Original post by AdeptDz)
    So would it be
      U_k+2 = 6(k-1)3k-9(k-2)3^{k-1} 

therefore

= 

6(k-1)3k-3^2(k-2)3^{k-1}   



therefore 

= 3[2(k-1)k-3(k-2)3^{k-2}]
    Rather than taking factorising 3 in the last step, do you notice that you'll get
    in the first term 3 \times 3^k and in the second term there is a 3^2 \times 3^k remembering that a^b \times a^c =a^{b+c}

    Also you've written it a bit wrong. It is
     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.
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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 \times 3^k and in the second term there is a 3^2 \times 3^k remembering that a^b \times a^c =a^{b+c}

    Also you've written it a bit wrong. It is
     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(3^{k+1}-3^k) - 3^{k+2} + 18 * 3^{k-1}
    • Community Assistant
    Offline

    3
    ReputationRep:
    (Original post by AdeptDz)
    cool so:
     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 \times 3^k \neq 3^{k+1}

     3^k \times 3^1 = 3^{k+1}

    So you want on the first term; 2[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=3^2

    But leave the brackets (k-1) and (k-2) for now.
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by NotNotBatman)
    The adding power rule only applies when it's the same base, so  k \times 3^k \neq 3^{k+1}

     3^k \times 3^1 = 3^{k+1}

    So you want on the first term; 2[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=3^2

    But leave the brackets (k-1) and (k-2) for now.
    oh my bad
    Would it be
     ..-3^2(k-2)3^{k-1}

    so 

  2[3^1(k-1)\times 3^{k}] - 3^2(k-2)3^{k-1}
    • Community Assistant
    Offline

    3
    ReputationRep:
    (Original post by AdeptDz)
    oh my bad
    Would it be
     ..-3^2(k-2)3^{k-1}]

    so 

  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.
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by AdeptDz)
    oh my bad
    Would it be
     ..-3^2(k-2)3^{k-1}

    so 

  2[3^1(k-1)\times 3^{k}] - 3^2(k-2)3^{k-1}
    so
    

2[3^{k+1}(k-1)] - 3^{k+1}(k-2)
    • Community Assistant
    Offline

    3
    ReputationRep:
    (Original post by AdeptDz)
    so
    

2[3^{k+1}(k-1)] - 3^{k+1}(k-2)
    Yes, then you can factorise.
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by NotNotBatman)
    Yes, then you can factorise.
     (3^{k+1})(k-1)-(k+2)
    Got a feeling i did that wrong as idk where to put the 2
    • Community Assistant
    Offline

    3
    ReputationRep:
    (Original post by AdeptDz)
     (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.

    3^{k+1}[2(k-1)-(k+2)] when you multiply out it's the same right?
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by AdeptDz)
    so
    

2[3^{k+1}(k-1)] - 3^{k+1}(k-2)
    (Original post by NotNotBatman)
    Yes, then you can factorise.
    1sec, let me have another go
    • Political Ambassador
    • Thread Starter
    Offline

    3
    ReputationRep:
    

2[3^{k+1}(k-1)] - 3^{k+1}(k-2)

2\times3^{k+1}(k-1) - 3^{k+1}(k-20)

3^{k+1}(2(k-1)-(k-2))

    there
 
 
 
Poll
Is GoT overrated?
Help with your A-levels

All the essentials

The adventure begins mug

Student life: what to expect

What it's really like going to uni

Rosette

Essay expert

Learn to write like a pro with our ultimate essay guide.

Uni match

Uni match

Our tool will help you find the perfect course for you

Study planner

Create a study plan

Get your head around what you need to do and when with the study planner tool.

Study planner

Resources by subject

Everything from mind maps to class notes.

Hands typing

Degrees without fees

Discover more about degree-level apprenticeships.

A student doing homework

Study tips from A* students

Students who got top grades in their A-levels share their secrets

Study help links and info

Can you help? Study help unanswered threadsRules and posting guidelines

Sponsored content:

HEAR

HEAR

Find out how a Higher Education Achievement Report can help you prove your achievements.

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.