(1/2+1/3+...+1/2006)(1+1/2+...+1/2005)-(1+1/2+...1/2006)(1/2+1/3+...1/2005)

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(1/2+1/3+...+1/2006)(1+1/2+...+1/2005)-(1+1/2+...1/2006)(1/2+1/3+...1/2005)
(1/2+1/3+...+1/2006)(1+1/2+...+1/2005)-(1+1/2+...1/2006)(1/2+1/3+...1/2005)

(1/2+1/3+...+1/2006)(1+1/2+...+1/2005)-(1+1/2+...1/2006)(1/2+1/3+...1/2005)
设(1+1/2+1/3+...+1/2005+1/2006)=x,则原式=(x-1)(x-1/2006)-x(x-1-1/2006) =x²-1/2006x-x+1/2006-x²+x+1/2006x =1/2006