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A= 1 x 3 + 3 x 5 + 5 x7 +...+ 95 x 97 + 97 x 99
\(\Rightarrow6A=1x3x6+3x5x6+5x7x6+...+95x97x6+97x99x6\)
\(6A=1x3x\left(5+1\right)+3x5x\left(7-1\right)+5x7x\left(9-3\right)+...+95x97x\left(99-93\right)+97x99x\left(101-95\right)\)
\(6A=1x3x5+1x3+3x5x7-1x3x5+5x7x9-3x5x7+...+95x97x99-93x95x97\)
\(+97x99x101-95x97x99\)
\(6A=\left(1x3x5+3x5x7+5x7x9+...+95x97x99+97x99x101\right)-\)
\(\left(1x3x5+3x5x7+...+93x95x97+95x97x99\right)+1x3\)
\(6A=97x99x101+1x3\)
\(6A=969903+3\)
\(6A=969906\)
\(A=\frac{969906}{6}\)
\(A=161651\)
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\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+111\right)}\)
\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{\left(1+1+...+1\right)+\left(2+2+...+2\right)+...+111}\)(\(111\)số hạng \(1\), \(110\)số hạng \(2\),...)
\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{1\times111+2\times110+3\times109+...+111\times1}\)
\(A=1\)
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Nè bé
B= 1.3+ 3.5 + 5.7 +... +97.99
2:B= 2/1.3 + 2/3.5 + 2/5.7 +.... + 2/97.99
2/B= 1/1-1/3+1/3-1/5+1/5-1/7+....+1/97-1/99
2/B=1/1-1/99
2:B=98/99
B=2: 98/99
B= 2. 99/98
B= 99/49
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Dấu " . " là dấu nhân
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)
\(A=\frac{49}{99}\)
\(B=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{97\times99}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(B=1-\frac{1}{99}\)
\(B=\frac{98}{99}\)
Số số hạng là :
(97-1):2+1=49
Tổng dãy là :
(97+1)(49:2)=2401
Đáp số : 2401
1+3+5+...+...+...+97
=(97+1)x[(97-1):2+1]:2
=2401