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\(B=3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+...+\frac{3}{1+2+3+4+...+100}\)
\(B=3.\left(\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+100\right).100:2}\right)\)
\(B=3.\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{100.101}\right)\)
\(B=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(B=6.\left(1-\frac{1}{101}\right)\)
\(B=6.\frac{100}{101}=\frac{600}{101}\)
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=98.99.100.101/4
=> A=24497550
1/1+(-2)+3+(-4)+.....+19+(-20)
=1-2+3-4+.....+19-20
=(1+3+.....+19)-(2+4+.....+20)
={(19+1).[(19-1):2+1]:2}-{(20+2).[(20-2):2+1]:2}
={20.10:2}-{22.10:2}
=10:2.(20-22)
=5.(-2)
=-10
a) \(1+2^1+2^2+2^3+....+2^{10}\)
\(\Rightarrow2A=2^1+2^2+2^3+....+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+....+2^{10}+2^{11}\right)-\left(1+2+2^2+2^3+....+2^{10}\right)\)
\(\Rightarrow A=2^{11}-1\)
b) \(3+3^2+3^3+3^4+.....+3^{100}\)
\(3A=3^2+3^3+3^4+....+3^{100}+3^{101}\)
\(3A-A=\left(3^2+3^3+3^4+....+3^{100}+3^{101}\right)-\left(3+3^2+3^3+....+3^{100}\right)\)
\(2A=3^{101}-3\)
\(A=\frac{3^{101}-3}{2}\)
c) \(2+2^2+2^3+....+2^{100}\)
\(2A=2^2+2^3+2^4+....+2^{100}+2^{101}\)
\(2A-A=\left(2^2+2^3+2^4+....+2^{100}+2^{101}\right)-\left(2+2^2+2^3+.....+2^{100}\right)\)
\(A=2^{101}-2\)