a:
\(1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\left(1\right)\)
Đặt \(S=1^2+2^2+...+n^2\)
Với n=1 thì \(S_1=1^2=1=\dfrac{1\left(1+1\right)\left(2\cdot1+1\right)}{6}\)
=>(1) đúng với n=1
Giả sử (1) đúng với n=k
=>\(S_k=1^2+2^2+3^2+...+k^2=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}\)
Ta sẽ cần chứng minh (1) đúng với n=k+1
Tức là \(S_{k+1}=\dfrac{\left(k+1+1\right)\cdot\left(k+1\right)\left(2\cdot\left(k+1\right)+1\right)}{6}\)
Khi n=k+1 thì \(S_{k+1}=1^2+2^2+...+k^2+\left(k+1\right)^2\)
\(=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(=\left(k+1\right)\left(\dfrac{k\left(2k+1\right)}{6}+k+1\right)\)
\(=\left(k+1\right)\cdot\dfrac{2k^2+k+6k+6}{6}\)
\(=\left(k+1\right)\cdot\dfrac{2k^2+3k+4k+6}{6}\)
\(=\dfrac{\left(k+1\right)\cdot\left[k\left(2k+3\right)+2\left(2k+3\right)\right]}{6}\)
\(=\dfrac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)
\(=\dfrac{\left(k+1\right)\left(k+1+1\right)\left[2\left(k+1\right)+1\right]}{6}\)
=>(1) đúng
=>ĐPCM
b: \(A=1\cdot5+2\cdot6+3\cdot7+...+2023\cdot2027\)
\(=1\left(1+4\right)+2\left(2+4\right)+3\left(3+4\right)+...+2023\left(2023+4\right)\)
\(=\left(1^2+2^2+3^2+...+2023^2\right)+4\left(1+2+2+...+2023\right)\)
\(=\dfrac{2023\cdot\left(2023+1\right)\left(2\cdot2023+1\right)}{6}+4\cdot\dfrac{2023\left(2023+1\right)}{2}\)
\(=\dfrac{2023\cdot2024\cdot4047}{6}+\dfrac{2023\cdot2024}{1}\)
\(=2023\left(\dfrac{2024\cdot4047}{6}+2024\right)⋮2023\)
\(A=\dfrac{2023\cdot2024\cdot4047}{6}+2023\cdot2024\)
\(=2024\left(2023\cdot\dfrac{4047}{6}+2023\right)\)
\(=23\cdot11\cdot8\cdot\left(2023\cdot\dfrac{4047}{6}+2023\right)\)
=>A chia hết cho 23 và 11
