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Xét tử ta có:
\(101+100+99+98+...........+3+2+1\)
\(=1+2+3+..........+99+100+101\)
\(=\frac{101.102}{2}=5151\)
Xét mẫu ta có:
\(101-100+99-98+.......+3-2+1\)
\(=\left(101-100\right)+\left(99-98\right)+.......+\left(3-2\right)+1\)
\(=1+1+.......+1+1=51\)
\(\Rightarrow A=\frac{5151}{51}=101\)
MỚI LÀM LÚC TỐI,HÊN QUÁ:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(4A=3-\left(\frac{101}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{203}{3^{100}}\)
\(A=\frac{3}{4}-\frac{203}{3^{100}\cdot4}< \frac{3}{4}\)
\(\frac{1}{2^1}+\frac{2}{3^2}+\frac{3}{4^3}+...+\frac{99}{100^{99}}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}<1\)
Vậy \(\frac{1}{2^1}+\frac{2}{3^2}+\frac{3}{4^3}+...+\frac{99}{100^{99}}<1\)
đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow A+3A=\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)+\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)<\(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow B+3B=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)+\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)\)
\(\Rightarrow4B=3-\frac{1}{3^{98}}<3\)
\(\Rightarrow B<\frac{3}{4}\Rightarrow4A<\frac{3}{4}\Rightarrow A<\frac{3}{16}\)
\(\RightarrowĐPCM\)
A=1.1+2.2+3.3+...+99.99+100.100
3A=1.2.3+2.3.(4-1)+...99.100.(101-98)
3A=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100
3A=99.100.101=999900
A= 999900:3=333300
\(A=1^2+2^2+3^2+...+100^2\)
\(\Rightarrow A=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)
\(\Rightarrow A=1.2-1+2.3-2+3.4-3+...+100.101-100\)
\(\Rightarrow A=1.2+2.3+...+100.101-\left(1+2+3+...+100\right)\)
\(\Rightarrow A=\frac{100.101.102}{3}-\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}=\frac{100.101.102.2}{6}-\frac{101.100.3}{6}\)
\(\Rightarrow A=\frac{100.101\left(102.2-3\right)}{6}\)
A=4+12+24+40+...+19404+19800
1/2A=2+6+12+...+9702+9900
1/2A=1.2+2.3+3.4+...+98.99+99.100
3/2A=1.2,3+2.3.(4-1)+3.4.(5-2)+...+98.99.(100-97)+99.100.(101-98)
3/2A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99+99.100.101-98.99.100
3/2A=99.100.101
A=(99.100.101):3/2=666600
B= 1+3+6+10+....+4851+4950
2B = 2+6+12+20+...+9702+9900
2B = 1.2+2.3+3.4+4.5+...+98.99+99.100
Xét A = 1.2+2.3+3.4+4.5+...+98.99+99.100
3A = 1.2.3+2.3(4-1)+3.4(5-2)+....+99.100(101-98)
3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+99.100.101-98.99.100
3A = 99.100.101
B = 333300
Thay A vào B ta được:
2B = 333300
B = 166650
MK chỉ làm được đến đây thôi
khó nhỉ từ xem đã
ghép 2 nhóm lại mà tính mình ko rõ đề