Correct Answer: Only I

Solution : I. $100^2-99^2+98^2-97^2+96^2-95^2+$ $94^2-93^2+\ldots \ldots+22^2-21^2$
$= (100-99)(100+99) + (98-97)(98+97) + .......... + (22-21)(22+21)$
$= (1)(199) + (1)(195) + ............. + (1)(43)$
Here, $a$ = 43 and $d$ = 4
Let the number of terms be $n$.
So, $a + (n-1)d = 199$
$⇒ 43 + 4(n-1) = 199$
$⇒ 4(n-1) = 156$
$⇒ n-1 = 39$
$⇒ n = 40$
Sum of AP = $\frac{n}{2}(2a + (n-1)d)=\frac{40}{2}(43 + 199)= 4840$
II. $\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
= $\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^2-\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
= $\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
= $\left(\mathrm{k}^8-\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
Hence, the correct answer is only I.