Sum of 20 terms of the series $–1^{2} + 2^{2} –3^{2} + 4^{2} – …$ is

Step 1: Group terms in pairs
$(-1^2 + 2^2) + (-3^2 + 4^2) + (-5^2 + 6^2) + \ldots$

Since 20 terms $\Rightarrow$ there are $\dfrac{20}{2} = 10$ pairs.

Step 2: Simplify each pair
Each pair is of the form $-(2k-1)^2 + (2k)^2$ where $k = 1, 2, 3, \ldots, 10$

$= (2k)^2 - (2k-1)^2$

Using $a^2 - b^2 = (a+b)(a-b)$:

$= (2k + 2k - 1)(2k - 2k + 1)$

$= (4k - 1)(1)$

$= 4k - 1$

Step 3: Sum all 10 pairs
$S = \displaystyle\sum_{k=1}^{10} (4k - 1)$

$= 4\displaystyle\sum_{k=1}^{10} k \ - \ \displaystyle\sum_{k=1}^{10} 1$

$= 4 \times \dfrac{10 \times 11}{2} - 10$

$= 4 \times 55 - 10$

$= 220 - 10$

$= 210$

Answer: Sum of 20 terms $= \boxed{210}$