chứng minh
C1:
\(1^2+2^2+3^2+...+\left(n-1\right)^2+n^2\)
\(=1.\left(2-1\right)+2.\left(3-1\right)+3\left(4-1\right)+...+\left(n-1\right)\left(n-1\right)+n\left(n+1-1\right)\)
\(=1.2-1+2.3-2+3.4-3+...+\left(n-1\right)n-\left(n-1\right)+n\left(n+1\right)-n\)
\(=\left(1.2+2.3+3.4+...+\left(n-1\right)n+n\left(n+1\right)\right)-\left(1+2+3+...+n\right)\)
\(=\dfrac{1}{3}.\left(1.2.3+2.3.3+3.4.3+...+\left(n-1\right)n.3+n\left(n+1\right).3\right)-\dfrac{\left(\dfrac{n-1}{1}+1\right)\left(n+1\right)}{2}\)
\(=\dfrac{1}{3}.\left(1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+\left(n-1\right)n.\left[\left(n+1\right)-\left(n-2\right)\right]+n\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]\right)-\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{1}{3}.\left(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+\left(n-1\right)n.\left(n+1\right)-\left(n-2\right)\left(n-1\right)n+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\right)-\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{1}{3}.n\left(n+1\right)\left(n+2\right)-\dfrac{n\left(n+1\right)}{2}\)
\(=n\left(n+1\right)\left(\dfrac{n+2}{3}-\dfrac{1}{2}\right)=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\left(đpcm\right)\)
C2:
- Khi \(n=1\) thì đẳng thức trở thành:
\(1^2=\dfrac{1.\left(1+1\right)\left(2.1+1\right)}{6}\left(đúng\right)\)
Giả sử đẳng thức đúng với \(n=k\) tức là:
\(1^2+2^2+3^2+...+\left(k-1\right)^2+k^2=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}\)
Ta chứng minh đẳng thức cũng đúng với \(n=k+1\) tức là:
\(1^2+2^2+3^2+...+k^2+\left(k+1\right)^2=\dfrac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)Thật vậy, theo giả thiết quy nạp, ta có:
\(1^2+2^2+3^2+...+\left(k-1\right)^2+k^2+\left(k+1\right)^2=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(=\left(k+1\right)\left[\dfrac{k\left(2k+1\right)}{6}+k+1\right]\)
\(=\left(k+1\right)\left(\dfrac{2k^2+k+6k+6}{6}\right)\)
\(=\left(k+1\right)\left(\dfrac{2k^2+7k+6}{6}\right)\)
\(=\left(k+1\right)\left(\dfrac{2k^2+4k+3k+6}{6}\right)\)
\(=\left(k+1\right)\left[\dfrac{2k\left(k+2\right)+3\left(k+2\right)}{6}\right]\)
\(=\dfrac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)
Vậy đẳng thức đúng với \(n=k+1\). Theo nguyên lí Quy nạp toán học, đẳng thức đúng với mọi số tự nhiên n khác 0.
