Tính B = 1 + 2+ 3+ ...+ 98+99B = 1 + (2 + 3 + 4+...+ 98 + 99).

Ta thấy tổng trong ngoặc gồm 98 số hạng, nếu chia thành các cặp ta có 49 cặp nên tổng đó là:

(2 + 99) + (3 + 98) +..+ (51 + 50) = 49.101 = 4949

khi đó B = 1 + 4949 = 4950

1+2+3+...+99

a Ta có 

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

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

=       0 +0+... +0 (25 cs 0)

=0 x25=0

a)B=0 

B=1.(2-1)+2.(3-1)+3.(4-1)+...+98.(99-1)+99.(100-1)=(1.2+2.3+3.4+...+98.99+99.100)-(1+2+3+...+98+99)

Đặt

C=1.2+2.3+3.4+...+98.99+99.100 và D=1+2+3+...+98+99

D là cấp số cộng bạn tự tính

3C=1.2.3+2.3.3+3.4.3+....+98.99.3+99.100.3=1.2.3+2.3.(4-1)+3.4.(5-2)+...+98.99.(100-97)+99.100.(101-98)

3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5+...--97.98.99+98.99.100-98.99.100+99.100.101=99.100.101 => C=33.100.101

1) 1 + 2 + 3 + 4 + ........ + 99

= 99 . (99 + 1) : 2

= 99 . 100 : 2

= 99.50 = 4950

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

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

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

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)}=\frac{1}{100}\)

A= 99/1+98/2+...+2/98+1/99

<=>A= (99/1-98)+(98/2+1)+....+(2/98+1)+(1/99+1)

<=>A= 100/100+100/2+...+100/98+100/99

A= 100( 1/100+1/2+...+1/98+1/99)

Vậy B=1/100

-----------------------Good luck-------------------

bai toan nay kho

Answer:

`B=1+3-5-7+9+11-....-397-399`

`=(1+3-5-7)+(9+11-13-15)+...+(393+395-397-399)`

`=(-8)+(-8)+...+(-8)`

`=(-8).100`

`=-800`

`C=1-2-3+4+5-6-7+...+97-98-99+100`

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

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

`=0`

\(D=2^{100}-2^{99}-2^{98}-...-2^2-2-1\)

\(\Rightarrow2D=2^{101}-2^{100}-2^{99}-...-2^3-2^2-2\)

\(\Rightarrow2D-D=\left(2^{101}-2^{100}-...-2^2-2\right)-\left(2^{100}-2^{99}-...-2-1\right)\)

\(\Rightarrow D=2^{101}-2^{100}-2^{100}+1\)

\(\Rightarrow D=2^{101}-2^{101}+1\)

\(\Rightarrow D=1\)

Bx3=1x3/1x2x3x4+1x3/2x3x4x5+...+1x3/97x98x99x100

Bx3=3/1x2x3x4+3/2x3x4x5+...+3/97x98x99x100

Bx3=1/1x2x3-1/2x3x4+1/2x3x4-1/3x4x5+...+1/97x98x99-1/98x99x100

BX3=1/1x2x3-1/98x99x100

BX3=1/6-1/970200

Bx3=161700/970200-1/970200

Bx3=161399/970200

B=161699/970200:3

B=161699/970200x1/3

B=161699/2910600

\(B=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+....+\frac{1}{97.98.99.100}\)

\(B=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{100-97}{97.98.99.100}\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)\)

\(B=\frac{1}{3}\cdot\frac{161699}{970200}=\frac{161699}{2910600}\)