Bạn ghi bị lộn đề rồi, hai số cuối phải là \(97.98.99\)và \(98.99.100\)

\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+97.98.99.4+98.99.100.4\)

\(=1.2.3.4+2.3.4\left(5-1\right)+3.4.5\left(6-2\right)+...+97.98.99\left(100-96\right)+98.99.100.\left(101-97\right)\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100\)

\(=98.99.100.101=97990200\Rightarrow A=\frac{97990200}{4}=24497550\)

1) Đặt \(A=1.2+2.3+3.4+....+98.99\)

Ta có:\(3A=3.\left(1.2+2.3+3.4+....+98.99\right)\)

\(3A=1.2.3+2.3.3+3.4.3+....+98.99.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+....+98.99.\left(100-97\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)

\(3A=98.99.100\Rightarrow A=\frac{98.99.100}{3}=323400\)

Ta có:\(\frac{A.y}{1}=184800\Rightarrow y=184800:323400=\frac{4}{7}\)

2)Đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\right).1428+185,8\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{37.38.39}\)

Tổng quát:\(\frac{2}{\left(a-1\right)a\left(a+1\right)}=\frac{1}{\left(a-1\right)a}-\frac{1}{a\left(a+1\right)}\)

Ta có:

\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{37.38.39}\)

\(2B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+...+\left(\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(2B=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\Rightarrow B=\frac{370}{741}:2=\frac{185}{741}\)

Khi đó \(A=\frac{185}{741}.1428+185,8=...........\) (tự tính ra)

(*)số ko đẹp mấy

\(D=\frac{30}{1.2.30}+\frac{30}{2.3.4}+\frac{30}{3.4.5}+...+\frac{30}{98.99.100}\)

\(=15.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(=15.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=15.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(=15.\frac{8249}{9900}=\frac{8249}{660}\)

\(D=\frac{30}{1.2.3}+\frac{30}{2.3.4}+\frac{30}{3.4.5}+...+\frac{30}{98.99.100}\)

\(=15\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(=15\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=15\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(=15.\frac{4949}{9900}=\frac{4949}{660}\)

Vậy \(D=\frac{4949}{660}\).

\(D=\frac{30}{1.2.3}+\frac{30}{2.3.4}+\frac{30}{3.4.5}+...+\frac{30}{98.99.100}\)

\(D=15.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(D=15.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(D=15.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(D=15.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(D=15.\frac{4949}{9900}\)

\(D=\frac{4949}{660}\)

A = 1.2 + 2.3 + ... + 99.100

3A = 1.2.3 + 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

3A = 999900

A = 333300

C = 1.2.3 + 2.3.4 + ... + 49.50.51

4C = 1.2.3.4 + 2.3.4.(4-1) + ... + 49.50.51.(52-48)

4c = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ... + 49.50.51.52 - 48.49.50.51

4C = 49.50.51.52

4C = 6497400

C = 1624350

Ta có :

a=1.2+2.3+3.4+...+99.100

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

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

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

3a=99.100.101

a=\(\frac{99.100.101}{3}\)

a=333300

Tính c làm tương tự

a = 1.2 + 2.3 + 3.4 + ... + 99.100

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

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

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

3a = 99 . 100 . 101

3a = 3 . 33 . 100 . 101

a = 33 . 100 . 101

a = 333300