2+2^2+2^3+...+2^100-2^101
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( 101+100+.......+3+2+1 ) / ( 101-100+100_99+........+ 4 - 3 + 2 - 1 )
= [ ( 101+1 )+( 100+2 )+....+( 52+50 )+ 51 ] / [ ( 101-100 )+(100-99)+........+( 4 - 3 )+( 2 - 1 )
= 102+102+.........+102+51 / 1+1+..............+1+1
= { [ 51( cặp) * 102 ] +51 } / [ 51(cặp) * 1 ]
= 5252 + 51 / 51
= 5253 / 51
= 103
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(3/2-2/2^2)*(4/3-2/3^2)*......*(101/100-2/100^2)
=( 6/2^2-2/2^2 )*( 12/3^2-2/3^2 )*.....*( 10100/100^2-2/100^2 )
= 4/2^2 . 10/3^2 .... 10098/100^2
= 1.4/2.2 . 2.5/3.3 .... 99.102/100.100
= ( 1.2.3.....99/2.3.4.....100 ) . ( 4.5.6.....102/2.3.4.....100 )
= 1/100 . 101.102/2.3
= 1/100 . 1717 = 1717/100
KL : ...
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A = \(\dfrac{101+100+98+97+...+3+2+1}{101-100+99-98+...+3-2+1}\)
= \(\dfrac{\left(101+1\right).101:2}{1+1+1+...+1}\)
= \(\dfrac{5151}{101}\) = 51
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\(\frac{2}{x.\left(x+2\right)}+\frac{2}{3.5}+\frac{2}{5.7}.+.....+\frac{2}{99.101}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x}=\frac{101}{101}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{1}\)
\(\Rightarrow x=1\)
Chữa đề: Ta thu gọn A
A=2+2^2+2^3+...+2^100-2^101
2A=2^2+2^3+2^4+...+2^101-2^102
2A-A=2^101-2-2^102
Vậy A= 2^101-2-2^102
\(\text{Đ}\text{ặt}B=2+2^2+2^3+...+2^{100}-2^{101}\)
\(\text{Đ}\text{ặt}A=2+2^2+...+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{101}-2^{102}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{101}\right)-\left(2+2^2+2^3+...+2^{100}\right)\)
\(A=2^2+2^3+2^4+...+2^{101}-2-2^2-2^3-...-2^{100}\)
\(A=2^{101}-2\)
Thay A vào B ta được:
\(B=2^{101}-2-2^{101}\)
\(B=2^{101}-2^{101}-2\)
\(B=0-2\)
\(B=\left(-2\right)\)