Sum of Floor of Square-Root of Integers!

$\large{\sum _{ i=1 }^{ n }{ \left\lfloor \sqrt { i } \right\rfloor } =\sum _{ i=1 }^{ n }{ i\left( { \left( i+1 \right) }^{ 2 }-i \right) } =\sum _{ i=1 }^{ n }{ 2{ i }^{ 2 }+i } }$

Now we keep the value of $1$ and keep going on. We will get upto $6$ as $203$ we see the difference is $14$ and our next $i$ is $7$ . So $7x=14$ or $x=2$ . This gives our value of $n$ as $50$