What is the sum of the squares of all two-digit numbers each of which is completely divisible by 4?

Correct option is C

Given:
We need to find the sum of the squares of all two-digit numbers that are divisible by 4.
Formula Used:

$$\sum_{i=1}^k i^2 = \frac{k(k + 1)(2k + 1)}{6}$$

n-th term of an arithmetic sequence:
$$a_n = a_1 + (n - 1)d$$
Solution:
The sequence is 12, 16, 20,....., 96  with a common difference d = 4 .​
$$a_n = 96, a_1 = 12$$, and d = 4:
96 = 12 + (n - 1)$$\times$$​4
84 = (n - 1)$$\times$$​ 4
n - 1 = 21

n = 22
Total numbers =  22
The sum of squares of the numbers = ​$$12^2 + 16^2 + 20^2 + \ldots + 96^2$$​​
This can be written as = ​$$(4 \times 3)^2 + (4 \times 4)^2 + (4 \times 5)^2 + \ldots + (4 \times 24)^2$$​​

= 1$$6 \times (3^2 + 4^2 + 5^2 + \ldots + 24^2)$$​
​$$\sum_{i=1}^{24} i^2 - \sum_{i=1}^2 i^2 = \frac{24 \times 25 \times 49}{6} - (1 + 4)$$​​
= 4900 - 5 = 4895
Now,  multiply by 16:
​$$16 \times 4895 = 78320$$​