D = 1.2 + 2.3+ 3.4 +...+ 99.100

=>3D=1.2.3+2.3.3+3.4.3+...+99.100.3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+99.100.(101-98)

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

=99.100.101-0.1.2

=99.100.101

=999900

=>D=999900:3=333300

Dn = 1.2 + 2.3 + 3.4 +...+ n (n +1)

=>3Dn=1.2.3+2.3.3+3.4.3+...+n(n+1).3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]

=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-2.3.4+....+n(n+1)(n+2)-(n-1)n(n+1)

=n.(n+1).(n+2)-0.1.2

=n.(n+1)(n+2)

=>Dn=n.(n+1)(n+2):3

 =>điều cần chứng minh

Câu a: Không hỏi nên không trả lời

Câu b:Gọi d là ƯCLN của n và n+1

Ta có: n chia hết cho d

n+1 chia hết cho d

=>(n+1)-n chia hết cho d

=>1 chia hết cho d

=>d=1

Vậy phân số n/n+1 là phân số tối giản

Câu c: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

=\(1-\frac{1}{50}\)

Vì: \(1-\frac{1}{50}\)<\(1\)

Vậy:\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)<\(1\)

a) \(S=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow S=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

b) \(S=1.2.3+2.3.4+...+n\left(n+1\right)\left(n+2\right)\)

\(4S=1.2.3.4+2.3.4.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(=n\left(n+1\right)\left(n+2\right)\left(n+2\right)\)

\(S=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)

c) \(S=1.4+2.5+3.6+...+n\left(n+3\right)\)

\(=1.2+1.2+2.3+2.2+3.4+3.2+...+n\left(n+1\right)+2n\)

\(=\left(1.2+2.3+3.4+...+n\left(n+1\right)\right)+2\left(1+2+3+...+n\right)\)

\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}+n\left(n+1\right)\)

\(=\frac{n\left(n+1\right)\left(n+5\right)}{3}\)

1. Đề thiếu

2. BĐT cần chứng minh tương đương:

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Ta có:

\(a^4+b^4+c^4\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\dfrac{1}{3}\left(ab+bc+ca\right)^2\ge\dfrac{1}{3}.3abc\left(a+b+c\right)\) (đpcm)

3.

Ta có:

\(\left(a^6+b^6+1\right)\left(1+1+1\right)\ge\left(a^3+b^3+1\right)^2\)

\(\Rightarrow VT\ge\dfrac{1}{\sqrt{3}}\left(a^3+b^3+1+b^3+c^3+1+c^3+a^3+1\right)\)

\(VT\ge\sqrt{3}+\dfrac{2}{\sqrt{3}}\left(a^3+b^3+c^3\right)\)

Lại có:

\(a^3+b^3+1\ge3ab\) ; \(b^3+c^3+1\ge3bc\) ; \(c^3+a^3+1\ge3ca\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

\(\Rightarrow VT\ge\sqrt{3}+\dfrac{6}{\sqrt{3}}=3\sqrt{3}\)

4.

Ta có:

\(a^3+1+1\ge3a\) ; \(b^3+1+1\ge3b\) ; \(c^3+1+1\ge3c\)

\(\Rightarrow a^3+b^3+c^3+6\ge3\left(a+b+c\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

5.

Ta có:

\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\) ; \(\dfrac{a}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{c}{b}}\) ; \(\dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\)

\(\Rightarrow\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}+\sqrt{\dfrac{a}{c}}\le\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=1\)

Câu 1:

\(VT=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(VT=1-\dfrac{1}{n}< 1\) (đpcm)

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 

3A = 1.2.( 3 + 0 ) + 2.3.( 4 - 1 ) + .. + 99.100.( 101 - 98 ) 

3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100 

3A = 99.100.101 

A = ( 99.100.101 ) : 3 = 333300 

Vậy A = 333300

mk làm câu b

A=1.2+2.3+3.4+.......+99.100

3.A =3.1.2+2.3.3+3.4.3+............+99.100.3

3.A= 1.2.3+2.3.(4-1)+3.4.(5-2) +..........+99.100.(101-98)

3.A=1.2.3+2.3.4-1.2.3 +3.4.5-2.3.4+............+99.100.101-98.99.100

vì cứ +2.3.4  lại -2.3.4 cứ như thế

3.A=99.100.101

A=(99.100.101):3

A=333300

chúc bạn may mắn trong học tập 

mk vừa học xong

a) Vì n.(n+1) = 1/n-1/n+1 suy ra n thuộc N      n khác 0

b) A=1/1*2+1/2*3+1/3*4+...+1/9.10

A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

A=1-1/10=9/10

Vậy A = 9/10