Correct Answer: Only I

Solution : Statement I: The value of $100^2-99^2+98^2-97^2+96^2-95^2+94^2$ $-93^2+\ldots \ldots+22^2-21^2$ is 4840.
Let us solve the series:
$100^2-99^2+98^2-97^2+96^2-95^2+94^2$ $-93^2+\ldots \ldots+22^2-21^2$
$= (100+99)(100-99) + (98+97)(98-97) + .... + (22+21)(22-21)$$= 100 + 99 + 98 + 97 +.....+21$---------------(1)
$=$ Sum of first 100 natural numbers - Sum of first 20 natural numbers
$= \frac{100×101}{2} - \frac{20×21}{2}$ [we know that the sum of n consecutive numbers is $\frac{n(n+1)}{2}$]
$= 4840$
Hence, the statement I is correct.
Statement II:
$\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
$=(k^2 +\frac{1}{k^2})(k-\frac{1}{k})(k+\frac{1}{k})(k^4 + \frac{1}{k^4})(k^4-\frac{1}{k^4})$
$=(k^2+\frac{1}{k^2})(k^2-\frac{1}{k^2})(k^8-\frac{1}{k^8})$
$=(k^4-\frac{1}{k^4})(k^8-\frac{1}{k^8})$
Which is not equal to $k^{16}-\frac{1}{k^{16}}$.
Hence, the correct answer is only I.