How to sum all odd numbers in range?

Question

How easy sum all odd numbers in range?

As instance: 25 + 27 + 29 + 31?

Which formula I should apply?

Any thoughts? Do you know how to sum the first $n$ odd numbers? — lulu, Jul 08 '19 at 12:34
Use https://math.stackexchange.com/questions/2107151/the-sum-of-the-first-n-odd-natural-numbers — lab bhattacharjee, Jul 08 '19 at 12:34
How to sum first I know. but I need sum range, not starting from 1 — Jony, Jul 08 '19 at 12:34
See here about arithmetic series; you should edit your question to include the background (from your comment) that you know how to sum the first $n$ odd numbers but not a series not starting from $1$ — J. W. Tanner, Jul 08 '19 at 13:00
Possible duplicate of The sum of the first n odd natural numbers — Chain Markov, Jul 08 '19 at 16:25
@YaniorWeg: in this question, unlike that question, OP is asking about sum of odd numbers $not$ starting with the first — J. W. Tanner, Jul 08 '19 at 16:56

score 1 · Accepted Answer · answered Jul 08 '19 at 12:42

1

Since you indicated in comments you know how to sum the first $n$ odd numbers ($n^2$):

$25+27+29+31=$

$(1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31)-$

$(1+3+5+7+9+11+13+15+17+19+21+23)=$

$ 16^2-12^2=(16+12)(16-12)=28\times4=112$

answered Jul 08 '19 at 12:42

J. W. Tanner

agrees with this formula for arithmetic series: average term ($28$ in this case) times number of terms ($4$ in this case) – J. W. Tanner Jul 08 '19 at 12:46
I need formula with N – Jony Jul 08 '19 at 12:47
using the approach illustrated in my answer, the sum of the odd numbers from $2M+1$ to $2N-1$ is $N^2-M^2=(N+M)(N-M)$ – J. W. Tanner Jul 08 '19 at 12:53
So, if I need sum between 25 and 31. it is, 25 27 29 31 I should do 15^2 - 13^2 ? – Jony Jul 08 '19 at 12:55
What is N and M? – Jony Jul 08 '19 at 12:55
$2M+1$ is the first term in the series and $2N-1$ is the last; in the example, $2M+1=25$ so $M=12,$ and $2N-1=31$ so $N=16$ – J. W. Tanner Jul 08 '19 at 12:57
is it included 25? Because it is 13 amount from 1 to 25 – Jony Jul 08 '19 at 13:12
Yes, $25$ is included; the $12$ terms $1...23$ are not – J. W. Tanner Jul 08 '19 at 13:19

score 0 · Answer 2 · answered Jul 08 '19 at 12:40

0

See that

$$25+27+29+31 = 25+(25+2) + (25+4) + (25 +6)$$

$$= 4(25) + 2+4+6 = 4(25) + 2(1+2+3).$$

answered Jul 08 '19 at 12:40

B. Goddard

2 Answers2