Question

G=1³+2³+3³+...+n³

Answer 1

Nhận xét: n³ - n = n(n² - 1) = (n - 1).n.(n + 1) => n³ = (n - 1).n.(n + 1) + n. Áp dụng ta có:

1³ = 0 + 1

2³ = 1.2.3 + 2

3³ = 2.3.4 + 3

....

n³ = (n-1).n.(n+1) + n

=> G = (1+ 2+3+...+n) + [ 1.2.3 + 2.3.4 + ...+ (n-1).n.(n+1) ]

Đặt B = 1+ 2+3+...+n

\(\Rightarrow B=\frac{\left(n+1\right).n}{2}\) (1)

Đặt C = 1.2.3 + 2.3.4 + ...+ (n-1).n.(n+1)

4.C = 1.2.3.4 + 2.3.4.(5 - 1) + ...+ (n-1).n.(n+1).(n+2 - n-2)

4.C = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ...+ (n-1).n.(n+1).(n+2) - (n-2).(n-1).n.(n+1) = (n-1).n.(n+1).(n+2)

=> 4.C = (n-1).n.(n+1).(n+2)

=> C = \(\frac{\text{(n-1).n.(n+1).(n+2)}}{4}\) (2)

Từ (1) và (2) \(\Rightarrow G=B+C=\frac{\left(n+1\right).n}{2}\frac{\text{(n-1).n.(n+1).(n+2)}}{4}\)

Answer 2

\(G=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2=\left(\frac{\left(n+1\right).n}{2}\right)^2=\frac{\left(n+1\right)^2.n^2}{4}\)

Answer 3

ta có x³=\(\left[\frac{\left(x+1\right)x}{2}\right]^2\) ^{-\(\left[\frac{\left(x-1\right)x}{2}\right]^2\)}

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

.......

n³=\(\left[\frac{\left(n+1\right)n}{2}\right]^2\)-\(\left[\frac{n\left(n-1\right)}{2}\right]^2\)

=> G=1³+2³+...+.n³=\(\left[\frac{n\left(n+1\right)}{2}\right]^2\)

 

 

Answer 4

Hỏi cái gì vậy? Tính tổng G hay tìm n?

Answer 5

tính tổng

Answer 6

ủa mà dag mũ 3 chuyển sang mũ 2 vậy