Let Me Derive The Formula First

Relevant wiki: Sum of n, n², or n³

Consider the ratio below:

$\begin{aligned} \frac{X}{Y} & = \frac{101+202+303+...+5050}{2^2+4^2+6^2+...+100^2} \\ & = \frac{\displaystyle 101 \sum_{k=1}^{50} k}{\displaystyle 4 \sum_{k=1}^{50} k^2} \\ & = \frac{101(50)(51)}{2} \times \frac{6}{4(50)(51)(101)} \\ & = \frac{3}{4} < 1 \end{aligned}$

$\implies \boxed{Y > X}$

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Clarification : From the relevant wiki , we know that $1+2+3+\cdots+n = \dfrac12 n(n+1)$ and $1^2+2^2+3^2+\cdots+n^2=\dfrac16 n(n+1)(2n+1)$ .

$X=101(1+2+3+4+5...+50$

$X=101\frac{(50)(51)}{2}=\boxed{128775}$

$Y=4(1^2+2^2+3^2+...+50^2)$

$Y=4(\frac{50(51)(101)}{6})=\boxed{171700}$

$\therefore Y>X$