a. Chứng minh quy nạp đẳng thức sau
$$1^2+2^2+3^2+...+n^2 = \frac{n(n+1)(2n+1)}{6}$$
Suy ra
$\lim \frac{1^2+2^2+3^2+...+n^2}{5n^3-n^2+1} = \frac16\lim \frac{n(n+1)(2n+1)}{5n^3-n^2+1}$
$=\frac16\lim \frac{\left ( 1+\frac1n \right )\left ( 2+\frac1n \right )}{5-\frac1n+\frac1{n^3}}=\frac{1}{15}$