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Let #a=1^2/1+2^2/3+3^2/5+....1001^2/2001, b=1^2/3+2^2/5+3^2/7+...+1001^2/2003# then #a-b=#?

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Jul 31, 2018

#a-b#

#=1^2/1+(2^2-1^2)/3+(3^2-2^2)/5+..... +(1001^2-1000^2)/2001-1001^2/2003#

#=1001-1001^2/2003=(1001*1002)/2003#

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