Ta co :

1/2! +2/3! +3/4! +... + 99/100!

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

=1 - 1/100! <1 

lik e nhe

Co Gai De Thuong

A = 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰

   = ( 2 + 2² + 2³ + 2⁴ + 2⁵ ) + ... + ( 2⁹⁶ + 2⁹⁷ + 2⁹⁸ + 2⁹⁹ + 2¹⁰⁰ )

   = 2 x ( 1 + 2 + 2² + 2³ + 2⁴ ) + ... + 2⁹⁶ x  ( 1 + 2 + 2² + 2³ + 2⁴ )

   = 2 x      31                          + ... +  2⁹⁶ x 31

   = 31 ( 2 + ... + 2⁹⁶ )

Vậy A chia hết cho 31       

A = 2 + 2² + 2³ + 2⁴ + 2⁵ + .... + 2⁹⁶ + 2⁹⁷ + 2⁹⁸ + 2⁹⁹ + 2¹⁰⁰

A = [2 + 2² + 2³ + 2⁴ + 2⁵] + ... + 2⁹⁵[2 + 2² + 2³ + 2⁴ + 2⁵]

A = 62 + ... + 2⁹⁵.62

A = 2.31 + .... + 2⁹⁵.2.31

A = 31.2.[2⁰ + 2⁵ + ... +2⁹⁵]

=> A \(⋮31\)

Ta có
\(A=2^1+2^2+2^3+...+2^{100}\)
\(A=\left(2^1+2^2+2^3+2^4+2^5\right)+....+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(A=2\left(1+2+2^2+2^3+2^4\right)+...+\left(1+2+2^2+2^3+2^4\right)\left(2^{96}+2^{96}+2^{96}+2^{96}\right)\)
\(A=\left(1+2^2+2^3+2^4+2^5\right)\left(2+...+2^{96}\right)\)
\(A=31.\left(2+...+2^{96}\right)⋮31\)

ai trả lời giúp tui đi

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

Ta có:
A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99

=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16

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