Tìm X (1.21+3.41+...+99.1001)X=512012+522012+...+992012+1002012
Ta thấy 1.21=1−21 ; 3.41=31−41;......;99.1001=991.1001
=> $(1-\frac{1}{2} + \frac{1}{3}-\frac{1}{4}+..........+\frac{1}{99}-\frac{1}{100}).X=
\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}$
=>(1+31+51+.....+991)−21−41−...−1001)X=512012+522012+...+992012+1002012
=> (1+31+51+.....+991)−(21+41+...+1001)X=512012+522012+...+992012+1002012
Ta tính ở : (1+31+51+.....+991)−(21+41+...+1001)
Ta cộng 21+41+...+1001) vào số bị trừ và số trừ.
Ta có: (1+31+51+.....+991)−(21+41+...+1001)
=(1+31+51+.....+991+21+41+...+1001)−2(21+41+...+1001)
=(1+21+31+.......+1001)−2(21+41+...+1001)
=(1+21+31+.......+1001(−(1+21+...+501)
=511+521+....+1001
=> (511+521+....+1001)X=512012+522012+...+992012+1002012
=> (511+521+....+1001)X=2012.(511+521+...+991+1001)
=> X=2012