For a positive integer n, simplify 12−22+32−42+⋯+(2n−1)2−(2n)2.
Solution — click to reveal
We can pair the terms and use the difference of squares factorization, to get (12−22)+(32−42)+⋯+[(2n−1)2−(2n)2]=(1−2)(1+2)+(3−4)(3+4)+⋯+[(2n−1)−(2n)][(2n−1)+(2n)]=(−1)(1+2)+(−1)(3+4)+⋯+(−1)[(2n−1)+(2n)]=−1−2−3−4−⋯−(2n−1)−2n=−22n(2n+1)=−2n2−n.