Compute the value of N=1002+992−982−972+962+⋯+42+32−22−12,where the additions and subtractions alternate in pairs.
Solution — click to reveal
Using the difference of squares factorization, we have N=(1002−982)+(992−972)+(962−942)+(952−932)+⋯+(42−22)+(32−12)=2(100+98)+2(99+97)+2(96+94)+2(95+93)+⋯+2(4+2)+2(3+1)=2(1+2+⋯+100)=2⋅2100⋅101=10100.