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Sửa : Chứng tỏ : \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{4949}{9900}\)
Ta có : \(VT=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{99.100-2}{2.99.100}\)
\(=\frac{4949}{9900}=VP\)
Study well ! >_<
A=1+3+6+10+...+4851+4950 2A
=2+6+12+20+...+9702+9900
2A=1.2+2.3+3.4+4.5+...+98.99+99.100
Xét B=1.2+2.3+3.4+4.5+...+98.99+99.100
3B=1.2.3+2.3(4−1)+3.4(5−2)+...+99.100(101−98)
3B=1.2.3+2.3.4−1.2.3+3.4.5−2.3.4+...+99.100.101−98.99.100
3B=99.100.101 B=333300
Thay B vào A ta được:
2A=333300
A=166650
nguồn:Câu hỏi của Nguyễn Nguyệt Minh - Toán lớp 6 - Học toán với OnlineMath
A=6+16+30+48+...+19600+19998
A : 2 = 3 + 8 + 15 + 24 + . . . + 9800 + 9999
A : 2 = 1.3 + 2.4 + 3.5 + 4.6 + . . . + 98.100 + 99.101
A : 2 = 1.[1+2] + 2.[1+3] + 3.[1+4] + 4.[1+5] + . . . + 98.[1+99] + 99.[1+100]
A : 2 = 1 + 1.2 + 2 + 2.3 + 3 + 3.4 + 4 + 4.5 + . . . + 98 + 98.99 + 99 + 99.100
A : 2 = 1 + 2 + 3 + 4 + . . . + 199 + 1.2 + 2.3 + 3.4 + 4.5 + . . . + 98.99 + 99.100
A : 2 = 4950 + 333300
A = 676500
nguồn:Câu hỏi của trinh thi quynh anh - Toán lớp 7 - Học toán với OnlineMath
M=1.2+2.3+3.4+......+2009.2010
3M=1.2.3+2.3.3+3.4.3+......+2009.2010.3
3M=1.2.3+2.3.(4-1)+3.4.(5-2)+......+2009.2010.(2011-2008)
3M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+2009.2010.2011
2008.2009.2010
3M=2009.2010.2011
M=(2009.2010.2011):3
M=8120598990:3
M=2706866330
Ta có:
\(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)
\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+98.99.100.4\)
\(\Rightarrow4A=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...98.99.100.101-97.98.99.100\)
\(\Rightarrow4A=98.99.100.101\)
\(\Rightarrow A=\dfrac{98.99.100.101}{4}\)
Vậy \(A=\dfrac{98.99.100.101}{4}\)
Ta có: \(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)
\(4A=\left(1.2.3+2.3.4+...+98.99.100\right)4\)
\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)...+98.99.100.\left(101-97\right)\)
\(4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100\)
\(4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+97.98.99.100-97.98.99.100+98.99.100.101\)
\(4A=98.99.100.101\)
\(\Rightarrow A=\dfrac{98.99.100.101}{4}=24497550\)
Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101\
4A=98.99.100.101
A=\(\dfrac{\text{98.99.100.101}}{4}\)
tick nha
Ta có: \(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)
\(4A=\left(1.2.3+2.3.4+...+98.99.100\right)4\)
\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)...+98.99.100.\left(101-97\right)\)
\(4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100\)
\(4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+97.98.99.100-97.98.99.100+98.99.100.101\)
\(4A=98.99.100.101\)
\(\Rightarrow A=\dfrac{98.99.100.101}{4}=24497550\)
A=6+16+30+48+...+19600+19998
2A = 1.3+2.4+3.5+...+99.101
B=2+5+9+14+...+4949+5049
2A = 1.4+2.5+3.6+...+99.102
C=1.2.3+2.3.4+3.4.5+...+98.99.100
4A = 1.2.3.4+2.3.4(5-1)+3.4.5.(6-2)+...+98.99.100.(101-97)
4A = 1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100
4A = 98.99.100.101
A=6+16+30+48+...+19600+19998
A : 2 = 3 + 8 + 15 + 24 + . . . + 9800 + 9999
A : 2 = 1.3 + 2.4 + 3.5 + 4.6 + . . . + 98.100 + 99.101
A : 2 = 1.[1+2] + 2.[1+3] + 3.[1+4] + 4.[1+5] + . . . + 98.[1+99] + 99.[1+100]
A : 2 = 1 + 1.2 + 2 + 2.3 + 3 + 3.4 + 4 + 4.5 + . . . + 98 + 98.99 + 99 + 99.100
A : 2 = 1 + 2 + 3 + 4 + . . . + 199 + 1.2 + 2.3 + 3.4 + 4.5 + . . . + 98.99 + 99.100
A : 2 = 4950 + 333300
A = 676500