Find the sum of the first tern terms of the series
\(\left( 1 \frac{3}{5} \right)^2 + \left( 2 \frac{2}{5} \right)^2 + \left( 3 \frac{1}{5} \right)^2 + 4^2 + \left( 4 \frac{4}{5} \right)^2 + \dots\)
+1224 There is probably a formula for this, but I don't know any other method other than brute force.
\((8/5)^2 + (12/5)^2 + (16/5)^2 + \dots + (44/5)^2\)
\((64/25) + (144/25) + \dots + (1936/25)\)
The sum from 64 to 1936 is \((1^2 + 2^2 + \dots 45^2) - (1^2 + 2^2 \dots + 7^2) = \frac{(45)(46)(91)}{6} - 140 = 31225\).
Now, we have \(\frac{31225}{25} = \boxed{1249}\)
