câu này ra 300 bạn ạ.Nhớ tick cho mk nha
Xét \(1^3+2^3+3^3+4^3+...+n^3\)
Ta thấy :
\(\left(n-1\right)n\left(n+1\right)\\ =\left(n^2-n\right)\left(n+1\right)\\ =n\left(n^2-n\right)+n^2-n\\ =n^3-n^2+n^2-n\\ =n^3-n\)
\(\Rightarrow n^3-n=\left(n-1\right)n\left(n+1\right)\)
\(\Rightarrow n^3=\left(n-1\right)n\left(n+1\right)+n\)
Ta có \(1^3+2^3+3^3+...+24^3=x^2\)
\(\rightarrow1^3=\left(1-1\right).1.\left(1+1\right)+1=1\)
\(\rightarrow2^3=\left(2-1\right).2.\left(2+1\right)+2=1.2.3+2\)
\(\rightarrow3^3=\left(3-1\right).3.\left(3+1\right)+3=2.3.4+3\)
............
\(\rightarrow24^3=\left(24-1\right).24.\left(24+1\right)+24=23.24.25+24\)
Ta có \(1^3+2^3+3^3+...+24^3=x^2\)
\(\Leftrightarrow1+1.2.3+2+2.3.4+3+...+23.24.25+24=x^2\)
\(\Leftrightarrow\left(1+2+3+...+24\right)+\left(1.2.3+2.3.4+...+23.24.25\right)=x^2\)
Đặt \(A=1+2+3+...+24\)
\(\Rightarrow A=\frac{\left(24+1\right).24}{2}=300\) (1)
Đặt \(B=1.2.3+2.3.4+...+23.24.25\)
\(4B=1.2.3.4+2.3.4.4+...+23.24.25.4\)
\(4B=1.2.3.4+2.3.4\left(5-1\right)+...+23.24.25.\left(26-22\right)\)
\(4B=1.2.3.4+2.3.4.5-1.2.3.4+...+23.24.25.26-22.23.24.25\)
\(4B=23.24.25.26\)
\(B=\frac{23.24.25.26}{4}\)
\(B=89700\) (2)
Từ (1) và (2)
\(\Rightarrow A+B=300+89700\)
\(\Rightarrow A+B=90000\)
Mà \(A+B=x^2\)
\(\Rightarrow x^2=90000\)
\(\Rightarrow x=\pm300\)
