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\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) tương tự
\(B=\frac{3^{101}+1}{4}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)
B = 1+ 2 + 3 + ... + 98 + 99
số số hạng từ 1 đến 99 là : (99 - 1) : 1 + 1 = 99
=) B = (99+1) . 99 : 2 = 4950
vậy B = 4950
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3B-B=1-\frac{1}{3^{99}}\)
\(B=\frac{1-\frac{1}{3^{99}}}{2}\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)
\(3A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
Chúc bạn học tốt ~
Đặt A=1.2+2.3+...+99.100
3A=1.2.3+2.3.3+...+99.100.3
=1.2.(3-0)+2.3(4-1)+....+99.100(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+...+99.100.101-98.99.100
=99.100.101-0.1.2
=99.100.101
=>\(A=\frac{99.100.101}{3}=333300\)
Đặt \(A=1.2+2.3+3.4+4.5+...+99.100\)\(\Rightarrow3.A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3\)\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.-2.3.4+4.5.6+3.4.5+...+\)\(99.100.101-98.99.100\)
\(=99.100.101\)
\(=999900\Rightarrow B=999900\div3=333300\)
Chưa chắc lắm đâu nha !
\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{98}{100}=\frac{49}{50}\)
C =\(\frac{1}{100}-\frac{1}{100.99}-...\)\(-\frac{1}{3.2}-\frac{1}{2.1}\)
C = \(\frac{1}{100}-\frac{1}{100}+\frac{1}{99}-\frac{1}{99}+...\)\(+\frac{1}{3}-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
C = 1