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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.

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  • $\begingroup$ Please use commas to differentiate between different terms of the series. $\endgroup$ Oct 29, 2015 at 17:19
  • $\begingroup$ Okay done.. can you help $\endgroup$ Oct 29, 2015 at 17:22
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    $\begingroup$ Do you know how to sum the terms $2,3,4,5,...,n$? $\endgroup$
    – copper.hat
    Oct 29, 2015 at 17:23
  • $\begingroup$ 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. $\endgroup$
    – David P
    Oct 29, 2015 at 17:24

2 Answers 2

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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, $$ S_n = \begin{cases} \frac{1}{4}(n^2 + 8n + 8) & \text{if $n$ is even},\\ \frac{1}{4}(n^2 + 8n + 7) & \text{if $n$ is odd}. \end{cases} $$

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  • $\begingroup$ We have $S_7 = \sum_{k=0}^7 u_k = 2 + 2 + \dots + 5 + 5= 28$. $\endgroup$ Oct 29, 2015 at 18:45
  • $\begingroup$ How do i find the $Sn$ if $1+1+2+2+3+3+.. $ added $1+1$ ? $\endgroup$ Oct 29, 2015 at 19:10
  • $\begingroup$ @GaneshPandey: Your question is not clear for me. Can you reformulate it ? $\endgroup$ Oct 29, 2015 at 20:08
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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.

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