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a) 3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] /3
b)
Nhân 4 vào hai vế ta được:
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] /3
b) 1-3+5-7+9-11+......+2005-2007
=(1-3)+(5-7)+(9-11)+.....+(2005-2007)
=(-2)+(-2)+(-2)+......+(-2)
=(-2).1004
=(-2008)
c) 1+2+3-4-5-6+7+8+9-10-11-12+...+97+98+99-100-101-102
=(1+2+3-4-5-6)+(7+8+9-10-11-12)+.....+(97+98+99-100-101-102)
=(-9)+(-9)+....+(-9)
=(-9).17
=(-153)
Xin lỗi nha 2 dòng cuối mk làm sai
b)1-3+5-7+9-11+......+2005-2007
=(1-3)+(5-7)+(9-11)+....+(2005-2007)
=(-2)+(-2)+(-2)+....+(-2)
=(-2).502
=(-1004)
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/ \( 1+2-3-4+...+97+98-99-100\)
\(= ( 1+2-3-4)+...+( 97+98-99-100)\)
\(= -4+...+( -4)\)
\(=(-4).25\)
\(=-100\)
a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)
b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
Answer:
\(A=1-2+3-4+5-6+...+99-100\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(99-100\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
Có cặp số:
\([\left(100-1\right):1+1]:2=50\) cặp
\(\Rightarrow A=\left(-1\right).50=-50\)
\(B=1+\left(-4\right)+2+\left(-5\right)+...+20+\left(-23\right)\)
\(=[1+\left(-4\right)]+[2+\left(-5\right)]+...+[20+\left(-23\right)]\)
\(=\left(-3\right)+\left(-3\right)+...+\left(-3\right)\)
Có cặp số:
\([\left(20-1\right):1+1]:2.2=20\) cặp
\(\Rightarrow B=\left(-3\right).20=-60\)