Question

[Nguyễn Văn Phú]

Bai14: Tinh

          A = 1.2²+2.3²+3.4²+...+98.99²

Answer 1

[Đoàn Đức Hà]

\(A=1.2^2+2.3^2+3.4^2+...+98.99^2\)

\(=\left(2-1\right).2^2+\left(3-1\right).3^2+\left(4-1\right).4^2+...+\left(99-1\right).99^2\)

\(=2^3-2^2+3^3-3^2+4^3-4^2+...+99^3-99^2\)

\(=\left(2^3+3^3+4^3+...+99^3\right)-\left(2^2+3^2+4^2+...+99^2\right)\)

\(B=2^3+3^3+4^3+...+99^3\)

Ta có công thức tính tổng: 

\(1^3+2^3+3^3+...+a^3=\left[\frac{a\left(a+1\right)}{2}\right]^2\)

Do đó \(B=\left(\frac{99.100}{2}\right)^2-1^3=24502499\)

\(C=2^2+3^2+4^2+...+99^2\)

Ta có công thức: 

\(1^2+2^2+3^2+...+a^2=\frac{a\left(a+1\right)\left(2a+1\right)}{6}\)

Suy ra \(C=\frac{99.100.199}{6}-1^2=328349\)

Suy ra \(A=B-C=24502499-328349=24174150\)