Câu hỏi của Cây bắp cải

Question

Chứng tỏ rằng : $\text{A}=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]$.

zZz Cool Kid_new zZz · Accepted Answer

Đặt $A_k=1+2+3+4+.....+k=\frac{k\left(k+1\right)}{2}\Rightarrow A_k^2=\frac{k^2\left(k+1\right)^2}{4}$$A_{k-1}=1+2+3+4+.....+\left(k-1\right)=\frac{k\left(k-1\right)}{2}\Rightarrow A_{k-1}^2=\frac{k^2\left(k-1\right)^2}{4}$$\Rightarrow A_k^2-A_{k-1}^2=\frac{k^2\left(k+1\right)^2-k^2\left(k-1\right)^2}{4}=\frac{k^2\left(k^2+2k+1-k^2+2k-1\right)}{4}=\frac{4k^3}{4}=k^3$Khi đó:$1^3=A_1^2$$2^3=A_2^2-A_1^2$$3^3=A_3^2-A_2^2$$.........................................................................................$$n^3=A_n^2-A_{n-1}^2$$\Rightarrow1^3+2^3+3^3+.....+n^3=A_n^2=\left(1+2+3+......+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2$Câu trả lời: Đặt $A_k=1+2+3+4+.....+k=\frac{k\left(k+1\right)}{2}\Rightarrow A_k^2=\frac{k^2\left(k+1\right)^2}{4}$$A_{k-1}=1+2+3+4+.....+\left(k-1\right)=\frac{k\left(k-1\right)}{2}\Rightarrow A_{k-1}^2=\frac{k^2\left(k-1\right)^2}{4}$$\Rightarrow A_k^2-A_{k-1}^2=\frac{k^2\left(k+1\right)^2-k^2\left(k-1\right)^2}{4}=\frac{k^2\left(k^2+2k+1-k^2+2k-1\right)}{4}=\frac{4k^3}{4}=k^3$Khi đó:$1^3=A_1^2$$2^3=A_2^2-A_1^2$$3^3=A_3^2-A_2^2$$.........................................................................................$$n^3=A_n^2-A_{n-1}^2$$\Rightarrow1^3+2^3+3^3+.....+n^3=A_n^2=\left(1+2+3+......+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2$

TRẦN ĐỨC VINH · Answer

Đề ghi sót . Vế cuối là móc vuông đó bình phương chưCâu trả lời: Đề ghi sót . Vế cuối là móc vuông đó bình phương chư

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)