Again the same!

$\dfrac{1^2}{1^2} + \dfrac{(1+2)^2}{2^2} + \cdots + \dfrac{(1+2+3+\cdots+16)^2}{16^2} = \, ?$ $\sum_{n=1}^{16}(\frac{\sum_{k=1}^{n}k}{n})^{2}= \displaystyle\sum_{n = 1}^{16}\frac{[\frac{n(n + 1)}{2}]^2}{n^2}=$
$\sum_{n = 1}^{16}\frac{(n + 1)^2}{4}= \sum_{n = 1}^{16}(\frac{n^2}{4}+\frac{n}{2}+\frac{1}{4})=$ $4+\frac{1}{2}\sum_{n = 1}^{16}n+\frac{1}{4}\sum_{n = 1}^{16}n^2=4+68+374=\Large{\color{#3D99F6}{\boxed{446}}}$

To make life easier, we have

$\displaystyle\sum_{n = 1}^{16}\frac{[\frac{n(n + 1)}{2}]^2}{n^2} = \sum_{n = 1}^{16}\frac{(n + 1)^2}{4}.$

It is not hard to evaluate the last summation since it is just equal to

$\displaystyle\frac{2^2 + 3^2 + 4^2 + ... + 16^2 + 17^2}{4} = \frac{\frac{17(18)(35)}{6} - 1}{4} = 446.$

(1+2)+2 = 1,5; (1+2+3)÷3 = 2; (1+2+3+4)÷4 = 2,5

AP: 1; 1,5; 2; 2,5; 3; 3,5; ...; (1+2+3...+16)÷16

an = a1 + (n-1) • r

a16 = 1 + (16-1) • 0,5

a16 = 8,5

1² + 1,5² + 2² + 2,5² + ... + 8,5² = 446