Đặt mẫu là A :
=> A = \(\frac{1}{99}\)+ \(\frac{2}{98}\)+ \(\frac{3}{97}\)+...+ \(\frac{98}{2}\)+ \(\frac{99}{1}\)
= ( \(\frac{1}{99}\)+ 1 ) + ( \(\frac{2}{98}\)+ 1 ) + ( \(\frac{3}{97}\)+ 1 ) +...+ ( \(\frac{98}{2}\)+ 1 ) + ( \(\frac{99}{1}\)+1 ) - 99 ( Vì ta đã cộng với 99 số 1 rồi nên phải trừ 99 )
= \(\frac{100}{99}\)+ \(\frac{100}{98}\)+ \(\frac{100}{97}\)+...+ \(\frac{100}{2}\)+ \(\frac{100}{1}\)- 99
= \(\frac{100}{99}\)+ \(\frac{100}{98}\)+ \(\frac{100}{97}\)+...+ \(\frac{100}{2}\)+ ( \(\frac{100}{1}\)- 99 )
= \(\frac{100}{99}\)+ \(\frac{100}{98}\)+ \(\frac{100}{97}\)+...+ \(\frac{100}{2}\)+ \(\frac{100}{100}\)
= 100.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+...+ \(\frac{1}{100}\))
=> Tử trên mẫu là :
\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+...+ \(\frac{1}{100}\)
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100.(\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+...+ \(\frac{1}{100}\))
=\(\frac{1}{100}\)
