Câu hỏi của Đặng Quốc Huy

Question

Tính S= 1^3+2^3+3^3+4^3+...................+100^3

Nguyen · Accepted Answer

Ta có:

$\left(1+2+...+k\right)^2=1^3+2^3+...+k^3$$\left(k\in N,k>0\right)$

C/m: $\left(1+2+...+k\right)^2=\left[\frac{k\left(k+1\right)}{2}\right]^2=\frac{k^2\left(k+1\right)^2}{4}$

$1^3=1^2$

$2^3=\left(1+2\right)^2-1^2$

...

$k^3=\left(1+2+...+k\right)^2-\left(1+2+...+k-1\right)^2$

$\Rightarrow$$\left(1+2+...+k\right)^2=1^3+2^3+...+k^3$

Áp dụng ta có:

$S=\left(1+2+...+100\right)^2=\left(5050\right)^2=25502500$Câu trả lời: Ta có:

$\left(1+2+...+k\right)^2=1^3+2^3+...+k^3$$\left(k\in N,k>0\right)$

C/m: $\left(1+2+...+k\right)^2=\left[\frac{k\left(k+1\right)}{2}\right]^2=\frac{k^2\left(k+1\right)^2}{4}$

$1^3=1^2$

$2^3=\left(1+2\right)^2-1^2$

...

$k^3=\left(1+2+...+k\right)^2-\left(1+2+...+k-1\right)^2$

$\Rightarrow$$\left(1+2+...+k\right)^2=1^3+2^3+...+k^3$

Áp dụng ta có:

$S=\left(1+2+...+100\right)^2=\left(5050\right)^2=25502500$

B.Thị Anh Thơ · Answer

Ta có (n-1)n(n+1)=n3 -n

=>n3 =n+(n1)n(n+1)

Áp dụng vào biểu thức A:

A=13 +23 +...+1003

=>A=1+2+1.2.3+3+2.3.4+...+100+99.100.101

=> A=(1+2+3+...+100) +(1.2.3+2.3.4+..+99.100.101)

=>A =5050+101989800

=> A=101994850Câu trả lời: Ta có (n-1)n(n+1)=n3 -n