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# FP2 Summation Question Watch

1. Show that

sum=(-1)+(1)-(1/2n+1)-(1/2n-1)

-1[(1/2n+1)+(1/2n-1)]
=-1[((2n-1)+(2n+1))/((2n-1)(2n+1))]
=-1[4n/(4n2-1)]
=4n/(1-4n2)

What am I doing wrong?
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2. No need to worry about this guys, I see my mistake.
This was part b of a question, for which in part a I had to find the partial fractions at the start of a question above. Just noticed now that the numerator is different.
3. (Original post by Nuclear Ghost)
Show that

sum=(-1)+(1)-(1/2n+1)-(1/2n-1)

-1[(1/2n+1)+(1/2n-1)]
=-1[((2n-1)+(2n+1))/((2n-1)(2n+1))]
=-1[4n/(4n2-1)]
=4n/(1-4n2)

What am I doing wrong?
Your answer is correct (I now see your question is wrong!).
4. (Original post by Nuclear Ghost)
Show that

sum=(-1)+(1)-(1/2n+1)-(1/2n-1)

-1[(1/2n+1)+(1/2n-1)]
=-1[((2n-1)+(2n+1))/((2n-1)(2n+1))]
=-1[4n/(4n2-1)]
=4n/(1-4n2)

What am I doing wrong?
Nothing. You are correct, it is the question which is wrong. Where did it come from?
5. (Original post by brianeverit)
Nothing. You are correct, it is the question which is wrong. Where did it come from?
Attached Images

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