Compute
123−2+133−3+143−4+⋯+11003−100
123−2+133−3+143−4+⋯+11003−100=100∑n=21n3−n=100∑n=21n(n2−1)=100∑n=21n(n−1)(n+1)=100∑n=21(n−1)n(n+1)=99∑n=11n(n+1)(n+2)=99∑n=11n(n+1)⋅1(n+2)|1n(n+1)=1n−1n+1=99∑n=1(1n−1n+1)⋅1(n+2)=99∑n=11n(n+2)−1(n+1)(n+2)|1n(n+2)=12(1n−1n+2)=99∑n=112(1n−1n+2)−1(n+1)(n+2)|1(n+1)(n+2)=1n+1−1n+2=99∑n=112(1n−1n+2)−(1n+1−1n+2)=99∑n=112n−12(n+2)−1n+1+1n+2=99∑n=112n−1n+1+1n+2(1−12)=99∑n=112n−1n+1+12(n+2)
123−2+133−3+143−4+⋯+11003−100=99∑n=112n−1n+1+12(n+2)=99∑n=112n−99∑n=11n+1+99∑n=112(n+2)=1299∑n=11n−99∑n=11n+1+1299∑n=11n+21299∑n=11n=12+14+1299∑n=31n=12+14+1297∑n=11n+299∑n=11n+1=12+99∑n=21n+1=12+98∑n=11n+2=12+97∑n=11n+2+11001299∑n=11n+2=1297∑n=11n+2+12⋅100+12⋅101=12+14+1297∑n=11n+2−(12+97∑n=11n+2+1100)+1297∑n=11n+2+12⋅100+12⋅101=12+14+1297∑n=11n+2−12−97∑n=11n+2−1100+1297∑n=11n+2+12⋅100+12⋅101=14−1100+12⋅100+12⋅101+1297∑n=11n+2−97∑n=11n+2+1297∑n=11n+2=14−1100+12⋅100+12⋅101+97∑n=11n+2−97∑n=11n+2=14−1100+12⋅100+12⋅101=12(12−150+1100+1101)
123−2+133−3+143−4+⋯+11003−100=12(12−150+1100+1101)=504920200=0.24995049505