A=2(1-3)+4(5-3)+ 6(5-7)+...+50(49-57)

A=-4-8-12-...-100 = -(4+8+12+...+100) (tính tổng cấp số cộng)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

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

Vậy A=49/50

Công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

A=1.2+2.3+...+49.50

3A=1.2.3+2.3.3+...+49.50.3

3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)

3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48

3A=49.50.51

=>A=49.25.51

=>A=62475

A=1.2+2.3+...+49.50

3A=1.2.3+2.3.3+...+49.50.3

3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)

3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48

3A=49.50.51

=>A=49.25.51

=>A=62475

3S=(1.2+2.3+3.4+...+49.50).3

3S=1.2.3+2.3.3+3.4.3+...+49.50.3

3S=1.2.3+2.3.(4-1)+3.4.(5-2)+...+49.50.(51-48)

3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50

3S=49.50.51

S=17.49.50

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

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

\(=1-\frac{1}{50}< 1\) (đpcm)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

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

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

\(\Rightarrow\) Quy đồng phân số và 1 là : \(\frac{49}{50}\) và \(1\)

Giữ nguyên phân số \(\frac{49}{50}\)

Ta có : \(\frac{1}{1}=\frac{1.50}{1.50}=\frac{50}{50}\)

\(\Rightarrow\frac{49}{50}< \frac{50}{50}\left(đpcm\right)\)

3.s =1.2.3+2.3.3 +3.4.3+........+49.50

3s= 1.2 .3+ 2.3.(4-1) +.....+ 49.50( 51-48)

3s=49.50.51

3s=124950

s= 124950chia 3

s= 41650

chia=/ or chia= :

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đặt A = 1.2. + 2.3 + 3.4 + ... + 49.50

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

3A = 49.50.51

A = 41650

Thay vào ta được

41650 + 1/2x = 40642

=> 1/2x = 1008

=> x = 2016

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

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

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

\(B=1.2+2.3+3.4+...+49.50\)

\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(=49.50.51\)

\(B=\frac{49.50.51}{3}=49.50.17\)

\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34

=> 3S = 32.33.34

=> S = \(\frac{32.33.34}{3}=11968\)