Question

Ch/m 1/2^2 +1/3^2 +1/4^2+...+1/100^2 <1

(BẠN NÀO TỐT GIÚP MÌN)

Answer 1

Ta có:1/2^2<1/1.2

1/3^2<1/2.3

1/4^2<1/3.4

....

1/100^2<1/99.100

Do đó 1/2^2+1/3^2+1/4^2+.....+1/100^2<1/1.2+1/2.3+1/3.4+...+1/99.100

=>1/2^2+1/3^2+1/4^2+........+1/100^2<1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100=1/1-1/100=99/100<1

=>1/2^2+1/3^2+1/4^2+....+1/100^2<1( đpcm)

Answer 2

ta có :
1/2^2+1/3^2+......+1/100^2<1/1.2+1/2.3+.....+1/99.100<1- 1/100<1(tớ làm chỗ này hơi tắt nhớ trình bày kĩ nha!)
Bài 4:
P<1+1+(1/1.2+1/2.3+....+1/1993.1994)(tính phần trong ngoặc sẽ bít nó nhỏ hơn 1
P<1+1+1=3

Answer 3

1/2^2+1/3^2+......+1/100^2
< 1/1.2+1/2.3+.....+1/99.100
< 1-1/2+1/2-1/3+.....+1/99-1/100
< 1- 1/100
< 99/100
=>1/2^2+1/3^2+......+1/100^2 < 99/100 < 1

Answer 4

S= 1/2^2+1/3^2+1/4^2+..............+1/100^2 < 1
S=1/2.2+1/3.3+1/4.4+..............+1/100.100
S<1/1.2+1/2.3+1/3.4+..............+1/99.100
S<1+1/2-1/2+1/3-1/3+1/4-1/4+...........+1/99-1/100
S<1-1/100
S<99/100<1
=> S<1

Answer 5

Answer 6

Ta có \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)

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

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

\(A=1-\frac{1}{2^{100}}<1\)

=> ĐPCM

 

Answer 7

Lê Minh Đức copy bá đạo