The sum of $n$ terms of the series $1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2}+......$

$\frac{n(n+3)}{4}$

The correct answer is Option (2) → $\frac{n(n+3)}{4}$

The given series is intended as:

$1^2 + 3^2 + 2^2 + 5^2 + 3^2 + 7^2 + \cdots$

i.e., it consists of pairs:

$(1^2 + 3^2), (2^2 + 5^2), (3^2 + 7^2), \ldots$

For the k-th pair:

$k^2 + (2k+1)^2$

The sum of the first n such pairs simplifies to:

$\frac{n(n+3)}{2}$