Answer:
See explanation
Step-by-step explanation:
Prove that
[tex]1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)[/tex]
1. When [tex]n=1,[/tex] we have
2. Assume that for all [tex]k[/tex] following equality is true
[tex]1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)[/tex]
3. Prove that for [tex]k+1[/tex] the following equality is true too.
[tex]1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)[/tex]
Consider left part:
[tex]1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}[/tex]
Consider right part:
[tex]\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)[/tex]
We get the same left and right parts, so the equality is true for [tex]k+1.[/tex]
By mathematical induction, this equality is true for all n.
