\(\frac{1}{2}+\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)( có 98 phân số => có 8 cặp )
\(=\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}=99.\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right)\)
\(\Rightarrow A=\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).1.2.3....98.99\)
\(\)A chia hết cho 99.
Vongola Primo
Ở đâu vậy bạn chỉ mình đi
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\left[2.\left(4.5.6...32\right)\left(34.35.36....98\right)\right]\left(3.33\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\left[2.\left(4.5.6...32\right)\left(34.35.36....98\right)\right].99\)
=>A chia hết 99 vì có thừa số 99
=>Đpcm
\(\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+..+\left(\frac{1}{49}+\frac{1}{50}\right)=\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{9}{49.50}\)
Ta gọi các thừa số phụ là: \(a_1;a_2;...;a_{49}\)
\(A=\frac{99.\left(a_1+a_2+...+a_{49}\right)}{2\times3\times....\times97\times98}.2.3.....97.98\)
\(A=99.\left(a_1+a_2+...+a_{49}\right)\)
\(\Rightarrow\) A chia hết cho 99.