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I was recently attempting to prove the formulae which calculate the sum of arithmetic sequences where the difference between each term is just 1.

I arrived at this formula first, which calculates the sum of the series from 1 to $n$, and then takes away values between 1 and $a$ - the starting term.

$\sum\limits_{a}^n r = \frac{(n+1)(n)}{2} - \frac{(a)(a-1)}{2}$

I later arrived at this formula:

$\sum\limits_{a}^n r = \frac{(n+a)(x)}{2}$

N.B. $x$ here denotes the number in the series that the last term is, rather than its value (e.g. 3rd, 4th etc.)

I however, up until now, have not found a way to make these formulae equal, even by attempting to substitute values for $a$ and $n$. Does anybody have a way to do this?

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  • $\begingroup$ you asked if $\frac{(n+1)n}{2}-\frac{a(a+1)}{2}=\frac{(n+a)n}{2}$? $\endgroup$ – Dr. Sonnhard Graubner Oct 23 '14 at 18:26
  • $\begingroup$ They are not equal. In the second way, the number of terms is not $n$. $\endgroup$ – André Nicolas Oct 23 '14 at 18:26
  • $\begingroup$ @AndréNicolas I have amended it thanks for pointing that out - before I was only working with series with the value of $a$ equal to 1. $\endgroup$ – Resquiens Oct 24 '14 at 19:35
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Hint: The second expression, $(n+a)n/2$, is not even an integer if you start with $a$ even and $n$ odd.

Edit: As suggested in the comments you most probably have not properly counted the number of terms from $a$ to $n$, which is $n-a+1$. So:

$$ \sum_{r=a}^n r = \frac{1}{2} \left( \sum_{r=a}^n r + \sum_{r=a}^n (n+a-r)\right) $$ $$ = \frac{1}{2} \sum_{r=a}^n (n+a) = \frac{1}{2} (n+a) \left(\sum_{r=a}^n 1\right) $$ $$= \frac{1}{2} (n+a)(n-a+1).$$

Edit2: Your original proof was: $$ \sum_{r=a}^n r = \sum_{r=1}^n r - \sum_{r=1}^{a-1} r$$ $$ = \frac{n(n+1)}{2} - \frac{(a-1)a}{2}$$ This is equal to the above expression as follows: $$ = \frac{n+n^2}{2} - \frac{(a-1)a}{2} $$ $$ = \frac{n+n^2-a^2+a}{2} = \frac{n+a + (n+a)(n-a)}{2}$$ $$ = \frac{(n+a)(n-a+1)}{2}.$$

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  • $\begingroup$ I just noticed thanks! The $n$ outside the brackets should really be $x$ where $x$ denotes the number in the series that the last term is, rather than its value... I've amended it. This is simply the normal formula for the summation of an arithmetic sequence, and I believe it will always be a whole number integer if the difference between each term is a whole number... By the way do you have any algebraic proof or insight as to why the sum of an arithmetic sequence using this is equation is always whole if the difference is also whole? $\endgroup$ – Resquiens Oct 24 '14 at 19:32
  • $\begingroup$ I'm not sure I understand the question. Are you asking why is $ \frac{1}{2} (n+a)(n-a+1)$ always an integer (whole)? $\endgroup$ – ir7 Oct 24 '14 at 19:49
  • $\begingroup$ Yes - that is it. $\endgroup$ – Resquiens Oct 24 '14 at 19:52
  • $\begingroup$ Basically, you can think of all combinations: ($a$ odd and $n$ even), ($a$ odd and $n$ odd) etc., and note that always one of the numbers $n+a$ or $n-a+1$ will turn out even, so their product is always even (multiple of 2). $\endgroup$ – ir7 Oct 24 '14 at 19:56
  • $\begingroup$ OK thanks - that explanation makes sense now! Can you explain your proof in the answer you have given above though? I'm not really sure what you've done? $\endgroup$ – Resquiens Oct 25 '14 at 13:42

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