Đáp án đúng: C
Ta có:
$\begin{array}{l}{{u}_{1}}=1\\{{u}_{2}}={{u}_{1}}+{{1}^{2}}\\{{u}_{3}}={{u}_{2}}+{{2}^{2}}={{u}_{1}}+{{1}^{2}}+{{2}^{2}}\\{{u}_{4}}={{u}_{3}}+{{3}^{2}}={{u}_{1}}+{{1}^{2}}+{{2}^{2}}+{{3}^{2}}\\....\\{{u}_{n}}={{u}_{{n-1}}}+{{(n-1)}^{2}}={{u}_{1}}+{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{(n-1)}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={{u}_{1}}+\frac{{(n-1).n.(2n-1)}}{6}\end{array}$
Chứng minh${{u}_{n}}={{u}_{1}}+\frac{{(n-1)n(2n-1)}}{6}$ bằng phương pháp quy nạp toán học.
Chọn C
