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

nth term of a sequence.... Watch

    • Thread Starter
    Offline

    2
    ReputationRep:
    [Edit]
    Offline

    0
    ReputationRep:
    Set n = 2k, and consider that -8 = (-2)^3. You should be able to make some progress from there (unless I've made a misunderstanding somewhere).
    Offline

    1
    ReputationRep:
    Yeah that sounds right.
    • Thread Starter
    Offline

    2
    ReputationRep:
    [Edit]
    Offline

    0
    ReputationRep:
    (Original post by flown_muse)
    Okay, now I have:

     a_2_k = \frac{(2k-1)(-2)^3^(^2^k^+^1^)}{2^(^2^k^-^1^)}

    and now I have an issue, because I can't find a value of k that can make both the top and the bottom equal to a power of 4.

    I'm clearly doing something wrong
    Write it as  a_2_k = (2k-1)\frac{(-2)^3^(^2^k^+^1^)}{2^(^2^k^-^1^)}, and remember that -2 = -1 \times 2, so you can expand the powers to get something easier.
    Offline

    1
    ReputationRep:
    (Original post by flown_muse)
    Okay, now I have:

     a_2_k = \frac{(2k-1)(-2)^3^(^2^k^+^1^)}{2^(^2^k^-^1^)}

    and now I have an issue, because I can't find a value of k that can make both the top and the bottom equal to a power of 4.

    I'm clearly doing something wrong
    Expand the numerator and simplify.
    • Thread Starter
    Offline

    2
    ReputationRep:
    [Edit]
    Offline

    1
    ReputationRep:
    (Original post by flown_muse)
     a_2_k = (2k-1)\frac{(-2)^3^(^2^k^+^1^)}{2^(^2^k^-^1^)}
     a_2_k = (2k-1)\frac{(-1)(2)^(^6^k^+^3^)}{2^(^2^k^-^1^)}
     a_2_k = (2k-1).(-1)(2)^(^6^k^+^3^)^-^(^2^k^-^1^)
     a_2_k = (2k-1).(-1)(2)^(^4^k^-^2^)

    etc.?
    What's 3--1?

    Apart from that yeah that's right. You can get rid of the 1 btw.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by f1mad)
    Yeah that's right. You can get rid of the 1 btw.
    So, I have ended up with:

     a_2_k = (2k-1).(-1)(2)^(^4^k^+^4^)
    a_2_k = (1-2k).2^4^k^+^4

    Therefore,  4k+4=4, 4k=0
    Offline

    1
    ReputationRep:
    (Original post by flown_muse)
    So, I have ended up with:

     a_2_k = (2k-1).(-1)(2)^(^4^k^+^4^)
    a_2_k = (1-2k).2^4^k^+^4

    Therefore,  4k+4=4, 4k=0
    Hang on, where did the last line come from?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by f1mad)
    Hang on, where did the last line come from?
    Because the question wants me to find An in terms of powers of 4, so I'm setting the power equal to 4.

    (I think?)
    • Thread Starter
    Offline

    2
    ReputationRep:
    Oh. I never thought of that. I think you are right, because otherwise I end up with n=0 and the answer is just a number with no powers. Argh! Thanks
    Offline

    1
    ReputationRep:
    (Original post by flown_muse)
    Oh. I never thought of that. I think you are right, because otherwise I end up with n=0 and the answer is just a number with no powers. Argh! Thanks
    Yeah you need to re-write it . In terms of 4^ something, since this would then imply powers of 4.
 
 
 
Poll
Do you honestly judge a book by its cover?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

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.