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    Can anyone solve this please? I don't know what to do for this question.

    Use method of PMI to prove that for all neZ+:

    (n+1) + (n+2) + (n+3) + ... + 2n = 1/2n(3n+1)

    Thanks
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    (Original post by sinnhy)
    Can anyone solve this please? I don't know what to do for this question.

    Use method of PMI to prove that for all neZ+:

    (n+1) + (n+2) + (n+3) + ... + 2n = 1/2n(3n+1)

    Thanks
    For all induction questions in FP1, there is a basic structure to follow.
    1. Clearly state the proposition you seek to prove.
    2. Test the statement in a base case i.e. set n=1 and show that LHS=RHS.
    3. Assume this statement is true for some value n=k, where k \in \mathbb{Z^+}.
    4. Consider the case n=k+1 and use your assumption for n=k to show that this is true. This is called the inductive step.
    5. Write a summarising statement to round off the proof i.e. explain how it follows by induction that this statement was true.
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    Sorry I didn't phrased it clearly. I've been stuck on the inductive step and can't seem to get the final bit for k+1th term which is suppose to be 1/2(k+1)(3k+4).

    I came up with the summation Sigma (n + r). and I added k+(k+1) to both sides? I dunno if i'm correct if anybody could help me.

    p.s I'm kinda new to studentroom, can anybody tell me how I can write up notations, maths equations etc in my post? thanks
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    Check what the k+1 term is and from this basically use the inductive step(P_k) where P(k) + k+1 = p(k+1).....rearrange and should be easy from there tbh!
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    Your neg rep is worth nothing OP and good lucking getting anyone else to help you with that kind of attitude!
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    (Original post by sinnhy)
    Sorry I didn't phrased it clearly. I've been stuck on the inductive step and can't seem to get the final bit for k+1th term which is suppose to be 1/2(k+1)(3k+4).

    I came up with the summation Sigma (n + r). and I added k+(k+1) to both sides? I dunno if i'm correct if anybody could help me.

    p.s I'm kinda new to studentroom, can anybody tell me how I can write up notations, maths equations etc in my post? thanks
    Consider ((k+1) + 1) + ..... ((k+1) + k) + ((k+1) + (k+1)) = (1/2)k(3k+1) + k + 2(k+1) which isn't quite what you said

    =(1/2)( 3k^2 + k +2k + 4k + 4 ) .... which gives you what you want?
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    (Original post by boromir9111)
    Your neg rep is worth nothing OP and good lucking getting anyone else to help you with that kind of attitude!
    I already know the fact, if you read my post above, I already tried that step and you was not helping much. I didn't mean to take offence.

    I managed to figure the question out, but thanks anyway for the help.
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    (Original post by ian.slater)
    Consider ((k+1) + 1) + ..... ((k+1) + k) + ((k+1) + (k+1)) = (1/2)k(3k+1) + k + 2(k+1) which isn't quite what you said

    =(1/2)( 3k^2 + k +2k + 4k + 2 ) .... which gives you what you want?
    That doesn't exactly factorise to the answer but thanks mate for trying
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    (Original post by sinnhy)
    That doesn't exactly factorise to the answer but thanks mate for trying
    Oops - sorry - now corrected - the last 2(k+1) expands to 2k + 2 which then becomes 4k + 4 when you put it over the common denominator of 2.
 
 
 
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