Collection of Maths Problems

Plus and minus

Task number: 3325

• Variant

  For all \(x\in S_n\): \(\displaystyle -\frac{n(n+1)}2\le x\le \frac{n(n+1)}2\).

• Solution

  Equality holds on the left when all signs are negative, so the sum can only be higher. Analogously for the right-hand side.

• Variant

  All numbers in \(S_n\) have the same parity (either all are even, or all are odd). How does this parity depend on \(n\)?

• Solution

  Adding an even term does not change the parity, while adding an odd term does change it. So as \(n\) increases the sums are as follows: odd, odd, even, even, odd, odd, even, even, etc.

• Variant

  All integers fulfilling the preceding two conditions lie in the subset \(S_n\).

• Solution

  We proceed by induction over the elements of \(S_n\) in ascending order.

  Base case of the induction: \(\displaystyle -\frac{n(n+1)}{2} = \sum_{i=1}^n -i\in S_n\).

  Let \(x\in S_n\) with \(x< \displaystyle \frac{n(n+1)}{2}\). Assume that \(x\) can be expressed as a sum \(\displaystyle \sum_{i=1}^n \pm i\) given an appropriate choice of signs.

  We will show that \(x+2\) belongs to \(S_n\). If the first term in the expression for \(x\) equals \(-1\), we replace it with \(+1\) and obtain an expression for \(x+2\).

  Otherwise because \(x< \displaystyle \frac{n(n+1)}{2}\), its sum must contain at least one term with a negative sign. Because the first term is positive (+1), the sum must contain two consecutive terms where the sign changes from positive to negative, i.e. \(+i-(i+1)\). We switch the signs of these two terms to \(-i+(i+1)\), which increases the entire sum by 2 and gives the expression we were seeking for \(x+2\).