The sum of n terms of the series

$1 + \frac{3}{2} + 2 + \frac{5}{2} + 3 + \frac{7}{2} + ...$

The series is:

$1 + \frac{3}{2} + 2 + \frac{5}{2} + 3 + \frac{7}{2} + ...$

Converting to fractions:

$\frac{2}{2} + \frac{3}{2} + \frac{4}{2} + \frac{5}{2} + \frac{6}{2} + \frac{7}{2} + ...$

The numerators are 2, 3, 4, 5, 6, 7... (consecutive numbers starting from 2)

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For the nth term:

• 1st term has numerator = 2 = (1+1)
• 2nd term has numerator = 3 = (2+1)
• 3rd term has numerator = 4 = (3+1)

General term: $T_n = \frac{n+1}{2}$

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The sum of n terms:

$S_n = \sum_{k=1}^{n} \frac{k+1}{2}$

$S_n = \frac{1}{2} \sum_{k=1}^{n} (k+1)$

$S_n = \frac{1}{2} \left[\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1\right]$

Using $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} 1 = n$:

$S_n = \frac{1}{2} \left[\frac{n(n+1)}{2} + n\right]$

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$S_n = \frac{1}{2} \left[\frac{n(n+1) + 2n}{2}\right]$

$S_n = \frac{1}{2} \left[\frac{n(n+1+2)}{2}\right]$

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

Therefore, the sum of n terms is $\frac{n(n+3)}{4}$