Question

[Đinh Tuấn Việt]

Chứng minh rằng với n \(\in\) N* thì :

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

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

Answer 1

[Trần Thị Loan]

a) A = 1² + 2² + ...+ n² = 1.(2 - 1) + 2.(3 - 1) + ...+ n.(n+ 1 - 1) = [1.2 + 2.3 + ...+ n.(n+1)] - (1 + 2 + ... + n)

Tính B = 1.2 + 2.3 + ...+ n.(n+1)

=> 3.B = 1.2.3 + 2.3.3 +3.4.3 + ...+ n.(n+1).3

= 1.2.3 + 2.3.(4 -1) + 3.4 .(5 - 2) + ...+ n.(n+1).((n+2) - (n-1) )

= [1.2.3.+ 2.3.4 + 3.4.5 +...+ n.(n+1).(n+2)] - [1.2.3 + 2.3.4 +...+ (n-1).n(n+1)] = n(n+1)(n+2)

=> B = n(n+1).(n+2)/3

Tính 1 + 2 + 3 + ..+ n =(n+1).n / 2

Vậy A =  n(n+1).(n+2)/3 - (n+1).n / 2 = n(n+1).(2n+1) / 6

Answer 2

[Trần Thị Loan]

Ta có: \(n^3=n.n.n=n.\left(\frac{n+1+n-1}{2}\right).n\left(\frac{\left(n+1\right)-\left(n-1\right)}{2}\right)\)

\(=\left(\frac{n\left(n+1\right)}{2}+\frac{n\left(n-1\right)}{2}\right).\left(\frac{n\left(n+1\right)}{2}-\frac{n\left(n-1\right)}{2}\right)=\left(\frac{n\left(n+1\right)}{2}\right)^2-\left(\frac{n\left(n-1\right)}{2}\right)^2\)

(Áp dụng công thức a² - b² = (a-b).(a+b))

Áp dụng vào ta có: \(1^3=\left(\frac{1.2}{2}\right)^2-\left(\frac{1.0}{2}\right)^2\)

                             \(2^3=\left(\frac{2.3}{2}\right)^2-\left(\frac{2.1}{2}\right)^2\)

                             \(3^3=\left(\frac{3.4}{2}\right)^2-\left(\frac{3.2}{2}\right)^2\)

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

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

Cộng từng vế ta được:

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