Question

Find the sum of the first 40 terms of series 1²+2²+3²+4²+.....16²

Answer 1

we know that a = 1
d= a2-a1=a3-a2
n = 40
we have use this formula
$sn = n\%2(2a + n - 1 \times d)$
➡️we get sum of the nth terms⬅️
✌️Hope this answer may help you ✌️
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Answer 2

HELLO........FRIENDS!!

THE ANSWER IS HERE,

$= > \: {1}^{2} + {2}^{2} + {3}^{2} ................$

Formula for the sum of squares of first n natural numbers is

$= > \: \frac{n(n + 1)(2n + 1)}{6}$

Here, no.of terms is 40.

So, n =40.

$= > \: \frac{40(41)(81)}{6}$

$= > \: 22140.$

:-)Hope it helps u.