S = 1/5 + 1/5² + 1/5³ + ... + 1/5¹⁰⁰

⇒ 5S = 1 + 1/5 + 1/5² + ... + 1/5⁹⁹

⇒ 4S = 5S - S

= (1 + 1/5 + 1/5² + ... + 1/5⁹⁹) - (1/5 + 1/5² + 1/5³ + ... + 1/5¹⁰⁰)

= 1 - 1/5¹⁰⁰

⇒ S = (1 - 1/5¹⁰⁰)/4

        S =           \(\dfrac{1}{5}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{5^3}\)+...+\(\dfrac{1}{5^{99}}\)+ \(\dfrac{1}{5^{100}}\)

      5S =     1 + \(\dfrac{1}{5}\) + \(\dfrac{1}{5^2}\) +  \(\dfrac{1}{5^3}\)+...+ \(\dfrac{1}{5^{99}}\)

5S - S =     1 - \(\dfrac{1}{5^{100}}\)

   4S   =       \(\dfrac{5^{100}-1}{4.5^{100}}\)

-50 mà bạn ơi bài nay lớp 7 hả

ngu thế

`S=1+2+3+4+5+...+100`

\(=\left(1+100\right)\times100:2\\ =101\times50=5050\)

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

=>S=(-1)+(-1)+(-1)+...+(-1) (có 50 số hạng -1)

=>S=(-1).50

=>S=-50

MS = 1/1 – 1/2 + 1/3 – 1/4 + ......... + 1/99 – 1/100

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

= (1 + 1/2 + 1/3 + 1/4 + ..... + 1/99 + 1/100) – (1/2 + 1/2 + 1/4 + 1/4 + 1/6 + 1/6 ....... 1/100 + 1/100) = (1 + 1/2 + 1/3 + 1/4 + ..... + 1/99 + 1/100) – (1 + 1/2  + 1/3 + ....... 1/50 ) = 1/51 + 1/52 + 1/53 + ............. + 1/100

Vậy TS/MS = 1

1-2+3-4+5-6+...+99-100+101 
= (1+3+5+...+101) - (2+4+6+...+100) 
tu 1 den 101 co : (101-1):2+1=51 
1+..+101 = (1+101)x 51:2= 2601 
tu 2 den 100 co : (100-2);2+1=50 
2+...+100 = (100 +2) x 50:2=2550 
=> S= 2601-2550=51

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

ta có: (100-1):1+1= 100

=>từ 1 đến 100 có 100 số

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

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

ta có: 100:2=50

=>có 50 số -1

=(-1).50

=-50

k mình nha bạn!