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. $\endgroup$– SchrodingersCatOct 29, 2015 at 17:19
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$\begingroup$ Okay done.. can you help $\endgroup$– Ganesh PandeyOct 29, 2015 at 17:22
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2$\begingroup$ Do you know how to sum the terms $2,3,4,5,...,n$? $\endgroup$– copper.hatOct 29, 2015 at 17:23
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$\begingroup$ 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 POct 29, 2015 at 17:24
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2 Answers
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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
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$\begingroup$ How do i find the $Sn$ if $1+1+2+2+3+3+.. $ added $1+1$ ? $\endgroup$ Oct 29, 2015 at 19:10
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$\begingroup$ @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.
answered Oct 29, 2015 at 17:27