\(\frac{1.4}{4.6}+\frac{2.5}{6.8}+...+\frac{48.51}{98.100}\)

=> \(\frac{1}{4}.\left(\frac{1.4}{2.3}+\frac{2.5}{3.4}+...+\frac{48.52}{49.50}\right)\)

=> \(\frac{1}{4}.\left(\frac{2.3-2}{2.3}+\frac{3.4-2}{3.4}+...+\frac{49.50-2}{49.50}\right)\)

=> \(\frac{1}{4}.\left(1-\frac{2}{2.3}+1-\frac{2}{3.4}+...+1-\frac{2}{49.50}\right)\)

=> \(\frac{1}{4}.\left[48-2.\left(\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{49.50}\right)\right]\)

=> \(\frac{1}{4}.\left[48-2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\right]\)

=> \(\frac{1}{4}.\left[48-2.\left(\frac{1}{2}-\frac{1}{50}\right)\right]\)

=> \(\frac{1}{4}.\left[48-2.\frac{12}{25}\right]\)

=> \(\frac{1}{4}.\frac{1176}{25}=\frac{249}{25}\)

\(Q=\frac{1}{4}\left(\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+...+\frac{48.51}{49.50}\right)\)

\(=\frac{1}{4}\left(\frac{2.3-2}{2.3}+\frac{3.4-2}{3.4}+\frac{4.5-2}{4.5}+...+\frac{49.50-2}{49.50}\right)\)

\(=\frac{1}{4}\left(1-\frac{2}{2.3}+1-\frac{2}{3.4}+1-\frac{2}{4.5}+...+1-\frac{2}{49.50}\right)\)

\(=\frac{1}{4}\left[48-2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\right]\)

\(=\frac{1}{4}\left[48-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\right]\)

\(=\frac{1}{4}\left[48-2\left(\frac{1}{2}-\frac{1}{50}\right)\right]=\frac{294}{25}\)

\(A = 1.4 + 2.5 + 3.6 + ...+ 99.102\)

\(A=1.2+1.2+2.3+2.2+3.4+3.2+...+99.100+99.2\)

\(A=(1.2+2.3+3.4+...+99.100)+2.(1+2+3+...+99)\)

\(A=333300+9900\)

\(A=343200\)

\(B = 2.4 + 4.6 + 6.8 + ....+ 98.100 + 100.102\)

\(B=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2) \)

\(B=4(1.2+2.3+...+50.51) \)

\(M= 1.2+2.3+...+50.51 \)

\(3M=1.2.3+2.3.(4-1)+...+50.51.(52-49) \)

\(=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51 \)

\(= 50.51.52\)

\(=132600 \)

\(\Rightarrow\)\(M=44200 \)

\(\Rightarrow\) \(B=4M=176800\)

Cảm ơn bạn

trong sách nâng cao và phát triển 6 đó bạn

a,6B=2.4.6+4.6.(8-2)+...............+98.100.(102-96)

6B=2.4.6+4.6.8-2.4.6+..............+98.100.102-96.98.100

6B=98.100.102

B=98.100.102:6

B=166600

Đặt A = 8.10 + 10.12 + 12.14 + ....... + 98.100

=> 6A = 8.10.12 - 8.10.12 + 10.12.14 - 10.12.14 + ...... + 98.100.102

=> 6A =  98.100.102

=> A = 98.100.102/6

=> A = 166600

c.1.2.3+2.3.4+4.5.6+5.6.7=6+24+120+210

                                      =30+120+210

                                      =150+210

                                      =360

E = 1.3 + 2.4 + 3.5 +...+ 97.99 + 98.100

A = 1.3 + 3.5 + 5.7 + ...+ 97.99

B = 2.4 + 4.6 + 6.8 + ... + 98.100

A = 1.3 + 3.5 + 5.7 + ... + 97.99

6A = 1.3.6 + 3.5.6 + 5.7.6 +...+ 97.99.6

1.3.6 = 1.3.(5+ 1) = 1.3.5 + 1.3.1

3.5.6 = 3.5(7 - 1) = 3.5.7 - 1.3.5

5.7.6 = 5.7.(9 - 3) = 5.7.9 - 3.5.7

7.9.6 = 7.9.(11 - 5) = 7.9.11 - 5.7.9

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

97.99.6 = 97.99.(101 - 95) = 97.99.101 - 95.97.99

Cộng vế với vế ta có:

6A = 1.3.1 + 97.99.101

6A = 3 + 969903

6A = 969906

A = 969906 : 6

A = 161651

B = 2.4 + 4.6 + 6.8 + ... + 98.100

6B = 2.4.6 + 4.6.6 + 6.8.6 + ... + 98.100.6

2.4.6 = 2.4.6

4.6.6 = 4.6.(8 - 2) = 4.6.8 - 2.4.6

6.8.6 = 6.8.(10 - 4) = 6.8.10 - 4.6.8

8.10.6 = 8.10.(12 - 6) = 8.10.12 - 6.8.10

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

98.100.6 = 98.100.(102 - 96) = 98.100.102 - 96.98.100

6B = 98.100.102

B = 98.100.102 : 6

B = 166600

E = A + B

E = 161651 + 166600

E = 328251

Ta có: \(A=1\cdot99+2\cdot98+3\cdot97+\cdots+98\cdot2+99\cdot1\)

\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)

\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\cdot\frac{49\cdot50}{2}-\frac{49\cdot\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)

\(=2\left\lbrack50\cdot49\cdot50-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)

\(=2\cdot\left\lbrack49\cdot50\cdot50-49\cdot25\cdot33\right\rbrack+2500\)

\(=2\cdot49\cdot25\cdot\left(2\cdot50-33\right)+2500\)

\(=49\cdot50\cdot67+2500=166650\)

Ta có: \(B=1\cdot2\cdot3+2\cdot3\cdot4+\ldots+17\cdot18\cdot19\)

\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+18\left(18-1\right)\left(18+1\right)\)

\(=2\cdot\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+18\left(18^2-1\right)\)

\(=\left(2^3+3^3+\cdots+18^3\right)-\left(2+3+\cdots+18\right)\)

\(=\left(1^3+2^3+\cdots+18^3\right)-\left(1+2+3+\cdots+18\right)\)

\(=\left(1+2+\cdots+18\right)^2-\left(1+2+\cdots+18\right)\)

\(=\left(18\cdot\frac{19}{2}\right)^2-18\cdot\frac{19}{2}=\left(9\cdot19\right)^2-9\cdot19=29070\)

Ta có: \(C=1\cdot4+2\cdot5+\cdots+100\cdot103\)

\(=1\left(1+3\right)+2\left(2+3\right)+\cdots+100\cdot\left(100+3\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)+3\left(1+2+\cdots+100\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{3\cdot100\cdot101}{2}\)

\(=\frac{100\cdot101\cdot201}{6}+\frac{3\cdot100\cdot101}{2}=50\cdot101\cdot67+3\cdot50\cdot101\)

\(=50\cdot101\cdot70=3500\cdot101=353500\)

Ta có: \(D=1\cdot3+2\cdot4+3\cdot5+\cdots+97\cdot99+98\cdot100\)

\(=1\left(1+2\right)+2\left(2+2\right)+3\left(3+2\right)+\cdots+97\cdot\left(97+2\right)+98\cdot\left(98+2\right)\)

\(=\left(1^2+2^2+\cdots+98^2\right)+2\cdot\left(1+2+3+\cdots+98\right)\)

\(=\frac{98\cdot\left(98+1\right)\left(2\cdot98+1\right)}{6}+2\cdot\frac{98\cdot99}{2}\)

\(=\frac{98\cdot99\cdot197}{6}+98\cdot99=49\cdot33\cdot197+98\cdot99=49\cdot33\left(197+2\cdot3\right)\)

\(=49\cdot33\cdot203=328251\)

a) \(A=2.4+4.6+...+98.100\)

\(\Rightarrow6A=2.4.6+4.6.6+....+98.100.6\)

\(=2.4.6+4.6.\left(8-2\right)+...+98.100.\left(102-96\right)\)

\(=2.4.6+4.6.8-2.4.6+...+98.100.102-98.98.100\)

\(=98.100.102\)

\(=999600\)

\(\Rightarrow A=\frac{999600}{6}=166600\)

PHẦN khác tương tự mẹo là xem tích đầu tiên rồi nhân cả biểu thức đó với số liền sau của tích các số đầu nhưng mà có quy luật