Advanced P13(BOMC 2)

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*Mahi*
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Advanced P13(BOMC 2)

Unread post by *Mahi* » Sun Apr 01, 2012 7:15 pm

Prove that for different choices of sign $+$ and $-$ the expression \[\pm 1 \pm 2 \pm 3 \pm \cdots \pm (4n+1)\] yields all odd positive integers less than or equal to $(2n+1)(4n+1)$
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*Mahi*
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Re: Advanced P13(BOMC 2)

Unread post by *Mahi* » Sun Apr 01, 2012 10:14 pm

Intuition:
Induction yields the most trivial solution.
Hint:
What I did:
Proved $\sum^{4n+1}_{i=1} i -2k=(4n+1)(2n+1)-2k\;$ ranges in all odds between $1$ and $(4n+1)(2n+1)$ where \[k=\sum^{4n+1}_{i=1} ix_i\] with all $x_i \in \{0,1\}$
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