1/2=1/2
1/3+1/4>1/4+1/4=1/2
1/5+…+1/8>4x1/8=1/2
1/9+…+1/16>8x1/16=1/2
1/2+1/3+1/4+…+1/16>4x1/2=2
1/2+1/3+1/4+…+1/63>1/2+1/3+1/4+…+1/16
suy ra: 1/2+1/3+1/4+…+1/63>2

hình như phân số cuối  phải là 1/324

nếu là 1/324 thì tớ giải nè:

A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324

= 1/4.(1+1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2+1/9^2)                                                    <1/4.(1+1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)

= 1/4.(1+1-1/9)

= 1/4.17/9 = 17/36<18/36 = 1/2

=> A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2

Cảm ơn nha Nobita Kun!!!

A<1/1*2+1/3*4+........+1/17*18

A<1-1/2+1/3-1/4+.......+1/17-1/18

A<(1+1/3+.....+1/17)-(1/2+1/4+......+1/18)

A<(1+1/2+1/3+......+1/18)-(1/2+1/4+.....+1/18)-(1/2+1/4+.......+1/18)

A<1-1/18-(1/2+1/3+1/4+......+1/17)

A<17/18-1/2-(1/3+1/4+......+1/17)

A<4/9-(1/3+1/4+.......+1/17)<1/2=4/8

Vậy a<1/2(đpcm)--------------------------Mình làm hơi dài nhé----------------------------------

a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}< 1\).

b. Có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\).

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}< 1\)

b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

   3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

3A-A=\(1-\frac{1}{3^{99}}\)

   2A=\(1-\frac{1}{3^{99}}\)

vì 2A<1

=> A<\(\frac{1}{2}\)

a. Nhân 2 vế của S với 3 rồi cộng S và 3S. Rút gọn sẽ ra kết quả

s=2+2^2+2^3+.....+2^100

s=2.(1+2+2^2+2^3)+......+2^97.(1+2+2^2+2^3)

s=2.15+....+2^97.15

s=15.(2+....+2^97)

=> s chia het cho 15

a=3+3^2+3^3+....+3^20

a=3.(1+3)+......+3^19.(1+3)

a=3.4+.....+3^19.4

a=4.(3+.....+3^19)

vay a chia het cho 4

tìm số tự nhiên n để a= 2n+6/n+1 là số nguyên tố