1/( 1+2+...+n)= 2/[n(n+ 1)]= 2[ 1/n- 1/(n+ 1)]
So the original formula
=2[ 1/2- 1/3]+2[ 1/3- 1/4]+...+2[ 1/99- 1/ 100]
=2[ 1/2- 1/3+ 1/3- 1/4+...+ 1/99- 1/ 100]
=2[ 1/2- 1/ 100]
= 1- 1/50
=49/50
High school final exam summary reflection 600 words 1
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