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

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.

• Please use commas to differentiate between different terms of the series. Oct 29, 2015 at 17:19
• Okay done.. can you help Oct 29, 2015 at 17:22
• Do you know how to sum the terms $2,3,4,5,...,n$? Oct 29, 2015 at 17:23
• the sequence is just $\{\lfloor(n+3)/2\rfloor\}$. You can find a formula for this by first considering $n$ odd and then $n$ even, look for a way to combine them. Oct 29, 2015 at 17:24

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, $$S_n = \begin{cases} \frac{1}{4}(n^2 + 8n + 8) & \text{if n is even},\\ \frac{1}{4}(n^2 + 8n + 7) & \text{if n is odd}. \end{cases}$$
• We have $S_7 = \sum_{k=0}^7 u_k = 2 + 2 + \dots + 5 + 5= 28$. Oct 29, 2015 at 18:45
• How do i find the $Sn$ if $1+1+2+2+3+3+..$ added $1+1$ ? Oct 29, 2015 at 19:10
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.