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A = 1*2+2*3+3*4+4*5+...+99*100
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 98.99.100 - 99.100.101
=> 3A = 99.100.101
=> A = 99.100.101 / 3 = 333300
S = 1 x 2 + 2 x 3 + ... + 99 x 100
3S = 1 x 2 x 3 + 2 x 3 x (4 - 1) + ..... + 99 x 100 x (101 - 98)
3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + .... + 99 x 100 x 101 - 98 x 99 x 100
3S = 99 x 100 x 101 = 999900
S = 999900 : 3 = 333300
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a) \(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{97.99}\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{3}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{3}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{32}{99}\)
\(=\frac{16}{33}\)
b)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}\)
\(=\frac{102}{103}\)
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đầu bài: 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20
= (1+19)+(2+18)+(3+17)+(4+16)+(5+15)+(6+14)+(7+13)+(8+12)+(9+11)+20
= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 +20
= 20.10
= 200
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20=\(\frac{\left(1+20\right)20}{2}\)=210
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a) \(\frac{15}{50}+20\%+\frac{5}{100}+0,45\)
\(=\frac{3}{10}+\frac{1}{5}+\frac{1}{50}+\frac{9}{20}\)
\(=\frac{97}{100}\)
b) \(\frac{3}{6}\times\frac{8}{12}\times\frac{15}{20}\times\frac{24}{30}\times\frac{35}{42}\times\frac{48}{56}\times\frac{65}{72}\times\frac{80}{90}\)
\(=(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7})\times\frac{65}{72}\times\frac{8}{9}\)
\(=\frac{1}{7}\times\frac{65}{72}\times\frac{8}{9}\)
\(=\frac{65}{567}\)
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\(y\div\frac{2}{10}+y\div12,5\%+4\div\frac{1}{y}+y+y\div0,5=\frac{20}{3}\)
\(\Rightarrow y\times5+y\times8+4\times y+y+y=\frac{20}{3}\)
\(\Rightarrow y\times\left(5+8+4+1+1\right)=\frac{20}{3}\)
\(\Rightarrow y\times19=\frac{20}{3}\)
\(\Rightarrow y=\frac{20}{3}\div19\)
\(\Rightarrow y=\frac{20}{57}\)
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\(M=1+2+3+4+5+...+99+100+101\)
Số số hạng của M là: \(\frac{101-1}{1}+1=101\)
Tổng của M = \(\frac{\left(1+101\right).101}{2}=5151\)
ĐS: 5151
M = 1 + 2 + 3 + 4 + 5 + ... + 99 + 100 + 101
Đây là dãy số cách đều 1 đơn vị
Số số hạng có trong dãy M là:
( 101 - 1 ) : 1 + 1 = 101 ( số hạng )
Tổng của dãy số M là:
( 101 + 1 ) x 101 : 2 = 5151
Đáp số: 5151
~ Chúc bạn học tốt ~
\(B=\frac{3}{5}+\frac{3}{10}+\frac{3}{20}+...+\frac{3}{100}\)
\(B=3\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{100}\right)\)
\(B=3\left(\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{20}+...+\frac{1}{50}-\frac{1}{100}\right)\)
\(B=3\left(\frac{1}{5}+\frac{1}{5}-\frac{1}{100}\right)\)
\(B=3\left(\frac{2}{5}-\frac{1}{100}\right)\)
\(B=3\cdot\frac{39}{100}\)
\(B=\frac{117}{100}\)
đáp án
117/100 = 1,17
hok tốt .