c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)

\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)

\(\Leftrightarrow A=33\cdot100\cdot101=333300\)

b) Ta có: \(1+2-3-4+...+97+98-99-100\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(=-4\cdot25=-100\)

Ta có :\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.....\frac{98^2}{98.99}=\frac{\left(1.2.3.4...98\right).\left(1.2.3.4...98\right)}{\left(1.2.3.4...98\right).\left(2.3.4.5...99\right)}=\frac{1}{99}\)

Lại có A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

Lại có \(A:B=\frac{98}{99}:\frac{1}{99}=98\)

=> A = 98B

các bạn có  về sweet home

C1 : B=\(\frac{1^2}{1.2}.\frac{2^2}{2.3}......\frac{98^2}{98.99}\)=\(\frac{1.1}{1.2}.\frac{2.2}{2.3}......\frac{98.98}{98.99}\)=\(\left(\frac{1.2......98}{1.2.....98}\right).\left(\frac{1.2......98}{2.3......99}\right)\)

                                                   \(1.\frac{1}{99}=\frac{1}{99}\)

C2:Đầu tiên cũng tách ra:\(1^2\)=1.1;\(2^2\)=2.2;...;\(98^2\)=98.98

Xong rút gọn ở tử và mẫu được:\(\frac{1}{2}.\frac{2}{3}.......\frac{98}{99}=\frac{1.2.....98}{2.3.....99}=\frac{1}{99}\)

Bạn thấy cách nào rễ hiểu hơn thì ghi nhé

a-b=(1.2+2.3+3.4+4.5+...+98.99)-(1²+2²+3²+...+98²)

1.2+2.3+3.4+4.5+...+98.99-1²-2²-3²-...-98²

^{=1(2-1)+2(3-2)+...+98(99-98)}

^=1+2+...+98

Đến đây bạn tự tính

b=1²+2²+3²+............+98²

b=1.(2-1)+2.(3-1)+3.(4-1)+..............+98.(99-1)

b=1.2-1+2.3-2+3.4-3+.............+98.99-98

b=(1.2+2.3+3.4+..................+98.99)-(1+2+3+............+98)

a-b=1+2+3+...............+98

a-b=\(\frac{98.\left(98+1\right)}{2}\)

a-b=4851

                                   Vậy A-B=4851

\(\text{a=1.2+2.3+3.4+......+98.99}\)

\(=1\left(1\text{+}1\right)\text{+}2\left(2\text{+}1\right)\text{+}3\left(3\text{+}1\right)\text{+}.........\text{+}98\left(98\text{+}1\right)\)

\(=1^2\text{+}1\text{+}1^2\text{+}2\text{+}3^2\text{+}3\text{+}...\text{+}98^2\text{+}98\)

\(=b\text{+}\left(1\text{+}2\text{+}3\text{+}...\text{+}98\right)\)

\(=b\text{+}\left(98\text{+}1\right).98:2\)

\(=b\text{+}4851\)

\(\Rightarrow a-b=4851\)

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