A=1+3/2^3+4/2^4+5/2^5+...100/2^100
1/2*A = 1/2 + 3/2^4 + 4/2^5 +....+ 99/2^100 + 100/2^101

A- A/2 = 1/2A =1/2 + 3/2^3 + 1/2^4 +...+1/2^100 - 100/2^101=

= [1/2+1/2^2 +1/2^3 +...+1/2^100] -100/2^101 (Do 3/2^3 = 1/2^2 +1/2^3)

=[1-(1/2)^101]/(1-1/2) -100/2^101 =

=(2^101 -1)/2^100 - 100/2^101

=> A= (2^101 -1)/2^99 - 100/2^100

minh biet lam ne nhung ban phai cho minh nhe

ai giup minh lam bai nay voi 

thanks nhieu

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

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

\(\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{100}{2^{101}}\)

\(\frac{1}{2}A-A=\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{100}{2^{101}}\right)\)

\(\frac{1}{2}A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}-\frac{1}{2}-\frac{3}{2^4}-\frac{4}{2^5}-...-\frac{100}{2^{101}}\)

\(\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)

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

\(\frac{1}{2}A=\frac{\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{101}\right]}{\frac{1}{2}}-\frac{100}{2^{101}}\)

A=2

\(B=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+.......+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)

\(\Leftrightarrow2B=1+\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+........+\dfrac{98}{2^{99}}+\dfrac{99}{2^{100}}\)

\(\Leftrightarrow2B-B=\left(1+\dfrac{1}{2^2}+\dfrac{2}{2^3}+........+\dfrac{99}{2^{100}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+......+\dfrac{100}{2^{100}}\right)\)

\(\Leftrightarrow B=\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{100}}-\dfrac{100}{2^{100}}\)

Đặt :

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{99}}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+......+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)

\(\Leftrightarrow B=1-\dfrac{1}{2^{100}}-\dfrac{100}{2^{100}}\)

\(\Leftrightarrow B=\dfrac{2^{100}-101}{2^{100}}\)

1/

Đặt A = 1+(-2)+3+(-4)+...+19+(-20)

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

Mỗi nhóm trên có số hạng là:

( 19-10):2+1 = 10 số hạng

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

A = 100 - 110

A = -10

2/

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

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

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

Từ 1 → 100 có 100 số hạng mà chia 2 số 1 nhóm

⇒ Số nhóm là:

100 : 2 = 50

mà mỗi nhóm bằng - 1

⇒ Tổng = - 50.

3/

a, 2-4+6-8+...+48-50

= ( 2-4)+( 6-8)+...+( 48-50)

= -2-2-...-2

= ( -2). 12

= -24

4/

-1+2-5+7-..+97-99

=(-1-99)+(-3-97)+...+(-49-51)

=(-100)+(-100)+...+(-100)

Có 50 cặp -100

Nên Tổng bằng : -100.50=-5000

Vậy....=-5000

5/

1+2-3-4+.....+97+98-99-100

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

=1+0+0+0+......+0+(-101)

=1+(-101)

=-100

Ta có : 1 + (-2) + 3 + (-4) + ...... + 19 + (-20)

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

= -1 + -1 + -1 + ..... + -1 

= -1.10

= -10