c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)

\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)

\(\Leftrightarrow A=33\cdot100\cdot101=333300\)

b) Ta có: \(1+2-3-4+...+97+98-99-100\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)

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

\(=-4\cdot25=-100\)

a) Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32

A = 1/2 + 1/2² + 1/2³ + 1/2⁴ + 1/2⁵

2A = 2(1/2 + 2² + 1/2³ + 1/2⁴ + 1/2⁵)

2A = 1 + 1/2 + 1/2² + 1/2³ + 1/2⁴

2A - A = (1 + 1/2 + 1/2²  + 1/2³ + 1/2⁴) - (1/2  + 1/2²  + 1/2³ + 1/2⁴ + 1/2⁵)

A = 1 - 1/2⁵

A = 31/32

b) 2/1.2 + 2/2.3 + 2/3.4 + ... + 2/18 . 19 + 2/19.20

= 2(1/1.2  + 1/2.3  + 1/3.4 + ... + 1/18.19 + 1/19.20)

= 2.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/18 - 1/19 + 1/19 - 1/20)

= 2. (1 - 1/20)

= 2.19/20

= 19/10

bạn thiếu ý C 

Sửa đề câu c) 

2 - 4 + 6 - 8 + 10 - 12 + ...... + 98 - 100 + 102 

= 2 + ( 6 - 4 ) + ( 10 - 8 ) + ..... + ( 98 - 96 ) + ( 102 - 100 )

= 2 + ( 2 + 2 + 2 + 2 + .... + 2 )

= 2 + 50

= 52

S = 1.2 + 2.3 + 3.4 +...+99.100

3S = 1.2.3 + 2.3.(4 - 1) + 3.4(5 - 2) +...+ 99.100(101 - 98)

3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ 99.100.101 - 98.99.100

3S = 99.100.101

3S = 999900

S = 333300

P = 1 + 3 + 5 + 7 +...+ 2015

P = (2015 + 1)1008 : 2 

P = 1016064

T = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 +...+ 97 + 98 - 99 - 100

T = (1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) +...+ (97 + 98 - 99 - 100)

T = (-4) + (-4) +...+ (-4)     

T = (-4)25

T = -100

S=999900

P=1016064

T=-100

S= 333300

quên chưa chia cho 3

a) 1 + 2 + 3 + 4 + ... + 100

= (100 + 1) x 100 : 2

= 5050

a) A=(100-1):1+1=100 số hạng   

    A=100:2=50 cặp

    tính giá trị của từng cặp số = (1+100)+(2+99)+(3+98)+...+(50+51)=101

    tính giá trị của biểu thức A: 50*101=5050

    [ mình tính theo công thức đó ]

K MIK NHA BẠN ^^

Tính B= 1 + 2 + 3 + ... + 98 + 99
Tính C = 1 + 3 + 5 + ... + 997 + 999
Tính D = 10 + 12 + 14 + ... + 994 + 996 + 998

4A=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

Bài 1: C = (999+1). [(999-1):2+1]: 2= 250000

Bài 2: B = (99+1). [(99-1):2+1]: 2= 2500

Bài 3: D = (998+10). [(998-10):2+1]: 2= 249480

Bài 4: 3S= 1.2.3 + 2.3.3 + 3.4.3+...+n.(n+1).3

              = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+n.(n+1).[(n+2)-(n-1)]

              = 1.2.3+2.3.4+2.3+3.4.5-2.3.4+.....+n.(n+1).(n+2)-n.(n+1)-(n-1)

              =n.(n+1).(n+2)

              => A = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Bài 1:
Số các số hạng trong tổng C là:
\(\left(999-1\right):2+1=500\)( số hạng)
=> \(C=\left(999+1\right).500:2=250000\)
Bài 2:
 Tổng B có số số hạng là: (99-1):1+1=99(số hạng)
=> \(B=\left(99+1\right)\times99:2=4950\)
Bài 3:
Số các số hạng trong tổng D là:
\(\left(998-10\right):2+1=495\)( số hạng)
=> \(D=\left(998+10\right).495:2=249480\)
Bài 4:
A = 1.2 + 2.3 + 3.4 + ... + n(n+1)
3A = 1.2.3 + 2.3.3 + 3.4.3+...+3n.(n+1)
3A = 1.2.3+2.3.(4-1)+3.4.(5-2)+...+n.(n+1){(n+2)-(n-1)}
3A = 1.2.3 + 2.3.4 -  1.2.3 +3.4.5 - 2.3.4 + .... + n(n+1)(n+2) - n(n+1)(n-1)
3A = n(n +1)(n+2)
=> A = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Vậy \(A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

A = chịu

B = ( 1 + 99 ) + ( 2 + 98 ) + ......

   = 100 . 50 = 5000

C = ( 1 + 999 ) + ( 3 + 997 ) + .....

   = 1000 . 500 = 500000

D = ( 10 + 998 ) + ( 12 + 996 ) + ......

   = 1008 . 495 = 498960

\(A=\frac{10}{1.2}+\frac{10}{2.3}+\frac{10}{3.4}+...+\frac{10}{98.99}+\frac{10}{99.100}\)

\(A=10.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=10.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=10.\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(A=10.\frac{99}{100}\)

\(A=\frac{99}{10}\)

Học tốt 

a= \(10.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

a=\(10.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

a=\(10.\left(1-\frac{1}{100}\right)\)

a=\(\frac{99}{10}\)

\(A=\frac{10}{1\cdot2}+\frac{10}{2\cdot3}+\frac{10}{3\cdot4}+...+\frac{10}{99\cdot100}\)

\(A=10\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\)

\(A=10\left(1-\frac{1}{100}\right)\)

\(A=10\cdot\frac{99}{100}\)

\(A=\frac{99}{10}\)

a/ Ta có :

\(10A=\frac{10\left(10^{50}+1\right)}{10^{51}+1}=\frac{10^{51}+10}{10^{51}+1}=\frac{10^{51}+1}{10^{51}+1}+\frac{9}{10^{51}+1}=1+\frac{9}{10^{51}+1}\)

\(10B=\frac{10\left(10^{51}+1\right)}{10^{52}+1}=\frac{10^{52}+10}{10^{52}+1}=\frac{10^{52}+1}{10^{52}+1}+\frac{9}{10^{52}+1}=1+\frac{9}{10^{52}+1}\)

Vì \(\frac{9}{10^{51}+1}>\frac{9}{10^{52}+1}\Leftrightarrow10A>10B\Leftrightarrow A>B\)

Vậy...

b/ Mình sửa lại một chút nhé :>

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}-3=0\)

\(\Leftrightarrow\left(\frac{x-1}{99}-1\right)+\left(\frac{x-2}{98}-1\right)+\left(\frac{x-3}{97}-1\right)=0\)

\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

\(\Leftrightarrow x-100=0\)

\(\Leftrightarrow x=100\)

Vậy...

c/ Đặt :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{1999.2000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{1999}-\frac{1}{2000}\)

\(=1-\frac{1}{2000}\)

\(=\frac{1999}{2000}\)

Vậy..