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# series

0
53
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Find the sum of the first tern terms of the series

$$\left( 1 \frac{3}{5} \right)^2 + \left( 2 \frac{2}{5} \right)^2 + \left( 3 \frac{1}{5} \right)^2 + 4^2 + \left( 4 \frac{4}{5} \right)^2 + \dots$$

Feb 19, 2021

#1
+944
+1

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 $$\frac{31225}{25} = \boxed{1249}$$

Feb 20, 2021