Nhân cả biểu thức với 3 nhé bn

Tử số = \(1.2.4+2.3.5+3.4.6+...+100.101.103\)

\(=1.2.\left(3+1\right)+2.3.\left(4+1\right)+3.4.\left(5+1\right)+...+100.101.\left(102+1\right)\)

\(=1.2.3+1.2+2.3.4+2.3+3.4.5+3.4+...+100.101.102+100.101\)

\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)+\left(1.2+2.3+3.4+...+100.101\right)\)

Mẫu số = \(1.2^2+2.3^2+3.4^2+...+100.101^2\)

\(=1.2.\left(3-1\right)+2.3.\left(4-1\right)+3.4.\left(5-1\right)+...+100.101.\left(102-1\right)\)

\(=1.2.3-1.2+2.3.4-2.3+3.4.5-3.4+...+100.101.102-100.101\)

\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)-\left(1.2+2.3+3.4+...+100.101\right)\)

đặt \(A=1.2.3+2.3.4+3.4.5+...+100.101.102\) và \(B=1.2+2.3+3.4+...+100.101\)

bạn tự tính : \(A=\frac{100.101.102.103}{4}=25.101.102.103\); \(B=\frac{100.101.102}{3}=100.101.34\)

rồi thay vào tìm P=\(\frac{A+B}{A-B}\)

A = 1.2.4 + 2.3.5 + ... + n(n+1)(n+3)

A = 1.2.(3+1) + 2.3.(4+1) + ... + n(n+1)[(n+2)+1]

A = [1.2.3 + 2.3.4 + ... + n(n+1)(n+2)] + [1.2 + 2.3 + ... + n(n+1)]

Đặt B = 1.2.3 + 2.3.4 + ... + n(n+1)(n+2)

4B = 1.2.3.(4-0) + 2.3.4.(5-1) + ... + n(n+1)(n+2)[(n+3)-(n-1)]

4B = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)

4B = n(n+1)(n+2)(n+3)

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

Đặt C = 1.2 + 2.3 + ... + n(n+1)

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

3C = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + ... + n(n+1)(n+2) - (n-1)n(n+1)

3C = n(n+1)(n+2)

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

A = B + C = \(n\left(n+1\right)\left(n+2\right)\left(\frac{n+3}{4}+\frac{1}{3}\right)\)

\(=n\left(n+1\right)\left(n+2\right)\frac{3n+13}{12}\)