Bài 3: 

a: =>4n-2-3 chia hết cho 2n-1

=>\(2n-1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{1;0;2;-1\right\}\)

b: =>-3 chia hết cho 2n-1

=>\(2n-1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{1;0;2;-1\right\}\)

A = 1/2.2 + 1/3.3 + ......+ 1/50.50

A < 1/1.2 + 1/2.3 +......+ 1/49.50

A < 1 - 1/2 + 1/2 - 1/3 +.......+ 1/49 - 1/50

A < 1 - 1/50

A < 49/50 < 1

=> A < 1 (đpcm)

*****k nha

Ta có: A=1/2^2+1/3^2+1/4^2+...+1/50^2<1

=> A<1/1.2+1/2.3+1/3.4+........+1/50.51

=>A< ( 1/1+ -1/2+1/2+ -1/3+1/3+ -1/4+1/4+ -1/5+1/5+.....+1/50+ -1/51)

=> A<1/1+ -1/51

=>A<51/51+ -1/51 =50/51<1

Ta có: 

1/2^2 < 1/1.2

1/3^2 < 1/2.3

......

1/50^2 < 1/49.50

=> A= 1/2^2+1/3^2+...+1/50^2 < 1/1.2+1/2.3+...+1/49.50 = 1-1/2+1/2-1/3+...+1/49-1/50 = 1-1/50<1 (ĐPCM)

Tính :(1-1/2)*(1-1/3)*(1-1/4)*...*(1-1/99)

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}

giai thich gio hon duoc ko

Ta có: A < \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

Lại có: \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

                                                                                 \(=1+\left(\frac{1}{1}-\frac{1}{50}\right)\)

                                                                                  \(=1+\frac{49}{50}\)

Mà 1+49/50<2 nên A<1+49/50<2

Vậy A<2

Ta có:
A=1+(1/2+1/3)+(1/4+1/5+1/6+1/7)+(1/8+1/9+......+1/15)+........+ (1/2^99+1/2^99+1+........+1/2^100-1)
(Có 99 nhóm) < 1+2.1/2+2^2.1/2^2+2^3.1/2^3+.....+2^99.1/2^99
=>1+1+1+.......+1 (100 số 1)=100
=>A1+1/2+2.1/2^2+2^2.1/2^3+2^3.1/2^4+.....+2^991/2^100-1-1/2^100 =1+1/2+1/2+1/2+1/2+........+1/2-1/2^100 (100 số 1/2)
=1+100.12-1/2^100
=50+1-1/2^100>50
=>A>50 (2)
Từ (1)và (2)=>50