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Question Number 11838 by Peter last updated on 02/Apr/17
22+122−1+32+132−1+42+142−1+....+202+1202−1=....?
Answered by ajfour last updated on 02/Apr/17
Tr=r2+1r2−1=r2−1+2r2−1=1+2(r−1)(r+1)Tr=1+1(r−1)−1(r+1)S=∑r=20r=2Tr=(1+11−13)+(1+12−14)+(1+13−15)+(1+14−16)+.......+(1+117−119)+(1+118−120)+(1+119−121)S=19+1+12−120−121S=20+169420.
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