The correct option is A 16n2+4n.
⇒S=4n2+(4n−1)2−(4n−2)2−(4n−3)2+...........−22−12→(II) ( in reverse order )
Adding them -
Factoring out (4n+1) all that's left is to complete the following sum ( let's call it A )
But, sum of each 4 consecutive numbers of A is equal to 8, because if we call the first number in each group of for x, then we have
As there are 4n numbers in the sum that is equal to A, there are n groups of 4 and ∴A=8n
Therefore, we have
Hence, the answer is 16n2+4n.