Ta có\(M=\left[\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\right].2.3...98\)
\(=\left[\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}\right].2.3...98=99\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).2.3...98\)
\(=99\left(\frac{k_1+k_2+...+k_{49}}{1.2.3...98}\right).2.3...98\left(k_1,k_2...k_{49}\varepsilonℕ^∗\right)=99\left(k_1+k_2+...+k_{49}\right)⋮99\Rightarrow M⋮99\left(đpcm\right)\)