Is there a general Formula for the series 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, $\dots n^{th}$ term 
Is there a general Formula for the series $2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7,\dots  n^{th}$ term

I want to calculate the sum upto $n$th term for the above series.
 A: We denote $(u_n)_{n\geq 0}$ the sequence. For all non-negative integer $n$, we have
$$
  u_{2n} = n + 2 \qquad\text{and}\qquad u_{2n+1} = n + 2.
$$
Let $S_n = \sum_{k=0}^n u_n$. Then
$$
  S_{2n + 1} = 2\sum_{k=2}^{n+2} k = 2(n+1)\times\frac{2 + (n+2)}{2} = n^2 + 5n + 4,
$$
(remenber that the sum of the terms of a finite arithmetical progression is obtained by multiplying the number of terms by the arithmetic mean of the first and the last terms)
and
$$
  S_{2n} = S_{2n+1} - (n+2) = n^2 + 4n + 2.
$$
If you prefer,
$$
  S_n =
    \begin{cases}
       \frac{1}{4}(n^2 + 8n + 8) & \text{if $n$ is even},\\
       \frac{1}{4}(n^2 + 8n + 7) & \text{if $n$ is odd}.
    \end{cases}
$$
A: If you have a sequence
$(a(n))_{n=1}^m$,
then
$(a(\lfloor \frac{n+1}{2} \rfloor)_{n=1}^m
$
repeats each term twice.
Note that
$(a(\lfloor \frac{n+k-1}{k} \rfloor)_{n=1}^m
$
repeats each term
$k$ times.
As for summing,
I recommend that you
consider even and odd $n$
separately.
I would get the
sum of an even number of terms first,
then the sum for an odd number of terms
is that sum plus the next term.
