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# Math Induction Question watch

1. 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
2. (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 and show that LHS=RHS.
3. Assume this statement is true for some value , where .
4. Consider the case and use your assumption for 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.
3. 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
4. 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!
5. Your neg rep is worth nothing OP and good lucking getting anyone else to help you with that kind of attitude!
6. (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?
7. (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.
8. (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
9. (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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