Is there a formula for (0.5)²+(1)²+(1.5)²+(2)²+...+(25)² (1/4)(1³) + (1/9)(1³+2³) + (16)(1³+2³+3³) + ... + (1/2704)(1³+2³+3³+...+51³) = ?
Initially, I factored it as 1³ (1/2² + 1/3² + ... + 1/52²) + 2³ (1/3² + ... + 1/52²) + 3³ (1/4² +...+ 1/52²) + ... + 51³ (1/52²), but got stuck to split them as separate series.
Alternatively, I observed this pattern
1³/4 + (1³/9 + 2³/9) + (1+8+27)/16 + 100/25 + ... + (1+2+3+...+51)/52
= (1/2)² + (3/3)² + (6/4)² + (10/5)² + ... + ((1+...+51)/52)²
= (1/2)² + (2/2)² + (3/2)² + (4/2)² + (5/2)² ... + (51/2)²
= (1/2)² (1+2²+3²+...+51²)
= (1/2)² (1/6)(51)(2(51)+1)(51+1)
= (1/2)² (1/6)(51)(103)(52)
= (1/2)² (17)(103)(26)
= 45526/4
= 11381.25
Does this seem correct?

The sum can be expressed as:
$$
\frac1{2^2}\cdot 1^3+\frac1{3^2}\cdot (1^3+2^3)+\frac1{4^2}\cdot (1^3+2^3+3^3)+\cdots +\frac1{52^2}\cdot (1^3+2^3+\cdots +51^3)=$$
$$\sum_{i=2}^{52} \frac1{i^2} \sum_{j=1}^{i-1} j^3\stackrel{(1)}{=}\\
\sum_{i=2}^{52} \frac1{i^2}\cdot (\frac{(i-1)i}{2})^2=\sum_{i=2}^{52} \frac{(i-1)^2}4=\frac14\sum_{i=1}^{51} i^2\stackrel{(2)}{=}\cdots$$
where:
$$\sum_{i=1}^{n} i^3=(\frac{n(n+1)}{2})^2   \quad  \quad  (1)$$
$$\sum_{i=1}^{n} i^2=\frac{n(n+1)(2n+1)}{6}   \quad  \quad  (2)$$
Can you finish?
