S(n)=1-2+3-4+5-...+(-1)^(n-1)n consists of two series S1(p)=1+3+5+7+...+(2p-1) and S2(q)=1+2+3+...+q, such that S(n)=S1(p)-2S2(q). S2(q)=(q+1)+((q-1)+2)+((q-2)+3)+...=q(q+1)/2 because the first and last, second and penultimate, and so on add up to the same number and there are q/2 such pairs when q is even; when q is odd the sum is q and there are q+1 pairs so the formula for the sum is q(q+1)/2, whether q is even or odd. We need to relate n, p and q. When n is even, say, 6, then the series is 1+3+5-2(1+2+3), so p=q=3=n/2; when n is odd, say, 7, the series is 1+3+5+7-2(1+2+3), so p=(n+1)/2 and q=(n-1)/2. S(n)=S1(n/2)-2S2(n/2) when n is even; S(n)=S1((n+1)/2)-2S2((n-1)/2) when n is odd.
The separation of the squares of the natural numbers is given by p^2-(p-1)^2=2p-1, so S1(p)=p^2, from the definition of S1 above.
S(71)=S1(36)-2S2(35). S2(35)=35*36/2=630. S1(36)=36^2=1296.
So, S(71)=1296-2*630=1296-1260=36.