S = 1/2 - 1/3 + 1/3 -1/4 + ......... +1/2011 -1/2012

S= 1/2 - 1/2012 = 1005/2012

\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...-\frac{1}{2012}\)

\(S=\frac{1}{2}+0+0+0+...-\frac{1}{2012}\)

\(S=\frac{1}{2}-\frac{1}{2012}\)

\(S=\frac{1005}{2012}\)

\(A=\frac{2012}{1}\cdot\frac{1005}{2012}\)

\(A=1005\)

\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(A=1-\frac{1}{2^9}\)

\(A=\frac{2^9-1}{2^9}\)

Vậy \(A=\frac{2^9-1}{2^9}\)

Chúc bạn học tốt ~ 

\(S=\frac{105}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+105}\)

\(=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\left(abc=105\right)\)

\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(bc+b+1\right)}\)

\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)

\(=\frac{bc+b+1}{bc+b+1}\)

\(=1\)

\(S=\frac{105}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+105}\)

\(=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)  \(\left(abc=105\right)\)

\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(bc+b+1\right)}\)

\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)

\(=\frac{bc+b+1}{bc+b+1}\)

\(=1\)

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)

Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ;   (1/51 + 1/52+...+1/59+1/60) > 1/6

S > 1/4 + 1/5 + 1/6.

Trong khi đó (1/4 + 1/5 + 1/6) > 3/5

=>S > 3/5                             (1)

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)

=> S <  4/5                             (2)

Từ (1) và (2) => 3/5 <S<4/5

A = 1/3.3/4.5/6...99/100

B = 2/3.4/5.6/7...100/101

Chứng minh A < B

Với: a; b; n ∈ N*; a < b ta có:

\(\frac{a}{b}\) = 1 - \(\frac{b-a}{b}\); \(\frac{a+n}{b+n}\) = 1 - \(\frac{b-a}{b+n}\)

Vì a < b nên b - a > 0

\(\frac{b-a}{b}\) > \(\frac{b-a}{b+n}\)

\(\frac{a}{b}\) < \(\frac{a+n}{b+n}\) (1) (hai phân số, phân số nào có phần bù nhỏ hơn thì phân số đó lớn hơn)

Áp dụng công thức (1) ta có:

\(\frac34\) < \(\frac{3+1}{4+1}=\frac45\)

\(\frac56<\frac{5+1}{6+1}=\frac67\)

.................................

\(\frac{99}{100}<\frac{99+1}{100+1}=\frac{100}{101}\)

Nhân vế với vế ta được:

3/4.5/6....99/100 < 4/5.6/7....100/101

suy ra:

A = 1/3.3/4.5/6....99/100 < 2/3.4/5.6/7..100/101 = B

A < B (Đpcm)

Câu b:

A = 1/3.3/4.5/6...99/100

B = 2/3.4/5.6/7...100/101

A.B = 1/3.3/4.5/6...99/100.2/3.4/5....100/101

A.B = \(\frac{1.3.5\ldots99}{3.5.7.\ldots101}\).\(\frac{2.4.6\ldots100}{3.4.6.\ldots100}\)

A.B = 1/101.2/3

A.B = 2/303