Cho:\(A=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right).2.3.4...98\)
Chứng minh rằng A chia hết cho 99
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)
Cho \(A=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right).2.3.4...98\)
Chứng minh A chia hết cho 99.
\(\frac{1}{2}+\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)( 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
Cho \(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right).2.3.4...98\). Chứng minh M chia hết cho 99
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
TRONG PHÉP NHÂN CÓ 3X33=99=>M LUÔN CHIA HẾT CHO 99
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{98}\right)\cdot2\cdot3\cdot4\cdot.........\cdot98\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{98}\right)\cdot\left(3\cdot33\right)\cdot2\cdot4\cdot......\cdot32\cdot34\cdot........\cdot98\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{98}\right)\cdot99\cdot2\cdot3\cdot.......\cdot32\cdot34\cdot........98\)
Vì \(99⋮99\Rightarrow M⋮99\)
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
\(=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+....+\left(\frac{1}{49}+\frac{1}{50}\right)=\frac{99}{1\times98}+\frac{99}{2\times97}+.....\frac{99}{49\times50}\)
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
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right).2.3.4...98\) Khi chia cho 99 có số dư là?
bạn dựa vào chuyên đề nâng cao đồng dư của lớp 6 ý. Lên mạng tra cũng có mà
A chia cho 99 không dư vì trong đẳng thức A có hai thừa số là 3 và 33 , ta có 3* 33=99 mà 99chia hết cho 99 nên A chia hết cho 99
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}\right)X2X3X4X...X98\)
Chứng minh rằng : M chia hết cho 99
Cho M=\(\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{98}\right).2.3.........98\)
Chứng tỏ rằng M chia hết cho 99
a,Cho A=\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\cdot2\cdot3\cdot4\cdot...\cdot98\)
CMR:A chia hết cho 99
b,Cho B=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{96}\) và B bằng phân số \(\frac{a}{b}\) .CMR A chia hết cho 97
Cho B=\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
Chứng minh B<1
\(B=\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+........+\frac{1^{99}}{2^{99}}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{99}}\right)\)
=>B=\(1-\frac{1}{2^{98}}\Rightarrow B
