1/ 1 + (-2) + 3 + (-4) + . . . + 19 + (-20)

=1-2+3-4+...+19-20

=(1-2)+(3-4)+...+(19-20)

=(-1)+(-1)+...+(-1)
=(-1).10

=-10

2/ 1 – 2 + 3 – 4 + . . . + 99 – 100

=(1-2)+(3-4)+...+(99-100)

=(-1)+(-1)+...+(-1)

=(-1).50

=-50

3/ 2 – 4 + 6 – 8 + . . . + 48 – 50

 =(2-4)+(6-8)+...+(48-50)

 =(-2)+(-2)+...+(-2)

 =(-2).13

 =-26

4/ – 1 + 3 – 5 + 7 - . . . . + 97 – 99

=(-1)+(3-5)+(7-9)+...+(97-99)

=(-1)+(-2)+(-2)+...+(-2)

=(-1)+(-2).45

=(-1)+(-90)

=(-91)

5/ 1 + 2 – 3 – 4 + . . . . + 97 + 98 – 99 - 100

=(1+2-3-4)+...+(97 + 98 – 99 - 100)

=(-4)+...+(-4)

=(-4).25

=-100

\(HT\)

1/ \(1+(-2)+3+(-4)+...+19+(-20)\)

\(=(-1+3+5+...+19)-(2+4+6+...+20)\)

\(=(19-1):2+1=10\)

\(=(1+19).10:2-(20+2).10:2\)

\(=100-110\)

\(=-10\)

2/ \(1 – 2 + 3 – 4 + . . . + 99 – 100\)

\(= ( 1 - 2 ) + ( 3 - 4) + .... + ( 99 - 100 )\)

\(= -1 + ( -1) + ....+ ( -1)\)

\(=(-1).50\)

\(=-50\)

3/ \( 2 – 4 + 6 – 8 + . . . + 48 – 50\)

\(= 2 +( – 4 + 6)+( – 8+10) + . . . +( -44+46)+ ( 48 – 50)\)

\(= 2+2+2+...+2+( -2) \)

\(= 2.12 +( -2 ) \)

\(=22\)

4/ \(-1+3-5+7-...+97-99\)

\(= ( -1 + 3 ) + ( -5 + 7 )+....+( -93 +95 ) + ( 97 - 99 )\)

\(= -2+( -2)+...+( -2)+2\)

\(= -2.24+2\)

\(=-46\)

5/ \( 1+2-3-4+...+97+98-99-100\)

\(= ( 1+2-3-4)+...+( 97+98-99-100)\)

\(= -4+...+( -4)\)

\(=(-4).25\)

\(=-100\)

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Ta có \(63,1.2-21,3.6=0,9.7.10.1,2-21.3,6\)

\(=6,3.1,2-21.3,6\)

\(=0,9.7.4.3-7.3.0,9.4\)

\(=6,3.1,2-6,3.1,2\)

\(=0\)

\(\Rightarrow\dfrac{\left(1+2+......+100\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}=\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+......+99-100}=0\)

\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...-\frac{100}{3^{100}}\)

\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow3A+A=1+\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{-2}{3^2}+\frac{3}{3^2}\right)+\left(\frac{3}{3^3}-\frac{4}{3^3}\right)+...+\left(\frac{-98}{3^{98}}+\frac{99}{3^{98}}\right)+\left(\frac{99}{3^{99}}-\frac{100}{3^{99}}\right)-\frac{100}{3^{100}}\)

\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow3.4A=3-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow3.4A+4A=3+\left(1-1\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{101}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow16A=3-\frac{99}{3^{99}}-\frac{100}{3^{100}}< 3\Rightarrow A< \frac{3}{16}< \frac{3}{4}\)

Lời giải:
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

\(3A=1-\frac{2}{3}+\frac{3}{3^2}-.....+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow 4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

$\Rightarrow 4A+12A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}<3$

$\Rightarrow 16A< 3$

$\Rightarrow A< \frac{3}{16}$

tham khảo: Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnlineMath

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Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnilneMath

1. 1-2+3-4+5-6-.....+99-100

=(1-2)+(3-4)+(5-6)+...+(99-100)                              (50 cặp)

=(-1)+(-1)+(-1)+...+(-1)                                          (50 số -1)

=(-1).50

=-50

2.1+3-5-7+9+11-.....-397-399

=(1+3-5-7)+(9+11-13-15)+....+(387+389-391-393)+395-397-399 (99 cặp)

=(-8)+(-8)+(-8)+...+(-8)+(-401)(có 99 có -8)

=(-8).99+(-401)

=(-792)+(-401)

=-1193

3. 1-2-3+4+5-6-7+...+96+97-98-99+100

=(1-2-3+4)+(5-6-7+8)+...+(93-94-95+96)+(97-98-99+100)            (25 cặp)

=0+0+0+...+0

=0

4. A=2¹⁰⁰-2⁹⁹-2⁹⁸-.....-2²-2-1

2A=2¹⁰¹-2¹⁰⁰-2⁹⁹-....-2³-2²-2

2A-A=A=2¹⁰¹-2¹⁰⁰-2¹⁰⁰+1

A=2¹⁰¹-2.2¹⁰⁰+1

A=2¹⁰¹-2¹⁰¹+1

A=1