cho A=\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right).2.3.4.5.....98\)
chứng minh A chia hết cho 99
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S
10 tháng 5 2017
Ta có: \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\)
\(=\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+\left(\frac{1}{3}+\frac{1}{96}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)
\(=\frac{99}{1.98}+\frac{99}{2.97}+\frac{99}{3.96}+...+\frac{99}{49.50}\)
\(=99\left(\frac{1}{1.98}+\frac{1}{2.97}+\frac{1}{3.96}+...+\frac{1}{49.50}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right).2.3.4....98\)
\(=99\left(\frac{1}{1.98}+\frac{1}{2.97}+\frac{1}{3.96}+...+\frac{1}{49.50}\right).2.3.4....98\)chia hết cho 99 (đpcm)
23 tháng 4 2016
\(\frac{1}{2}+\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)( 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.
29 tháng 4 2015
Ta có : M= [(1+1/98)+(1/2+1/97)+...+(1/49+1/50)].2.3.4...98
M=(99/1.98+99/2.97+...+99/49.50).2.3.4...98
M=99(1/1.98+1/2.97+...+1/49.50).2.3.4...98
M=99(k1+k2+...+k49/1.2.3.4...97.98).2.3.4...98
M=99(k1+k2+...+k49)
Vậy M chia hết cho 99
HT
1
25 tháng 4 2015
Ta có: \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}\right)=\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)
\(=\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}=99.\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right)\)
\(=9.11\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right)\)
Vậy: đpcm
TQ
10 tháng 10 2015
\(B=\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+........+\frac{1^{99}}{2^{99}}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{99}}\right)\)
=>B=\(1-\frac{1}{2^{98}}\Rightarrow B
\(=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+....+\left(\frac{1}{49}+\frac{1}{50}\right)=\frac{99}{1\times98}+\frac{99}{2\times97}+.....\frac{99}{49\times50}\)
Ta gọi các thừa số phụ là : \(a_1,a_2,......,a_{49}\)
\(A=\frac{99\times\left(a_1+a_2+.....+a_{49}\right)}{2\times3\times......\times97\times98}\times2\times3\times......\times97\times98\)
\(A=99\times\left(a_1+a_2+.....+a_{49}\right)\)
\(\Rightarrow A:99\)
\(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)(Có 98 phân số => có 49 cặp)
\(=\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}=99.\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right)\)
=> \(A=\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).1.2.3...98.99\)
=> A : 99 = \(\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).1.2.3...98=2.3.4...97+1.3.4..96.98+...+1.2.3..48.51...98\)
kết quả là số tự nhiên
=> A chia hết cho 99