What is the sum of (n-1)+(n-2)+...+(n-k)?

Question

What is the sum of this series ?

$(n-1)+(n-2)+(n-3)+...+(n-k)$

$(n-1)+(n-2)+...+3+2+1 = \frac{n(n-1)}{2}$

So how can we find the sum from $n-1$ to $n-k$ ?

score 13 · Answer 1 · edited Nov 08 '13 at 16:32

13

$$(n-1)+(n-2)\cdots(n-k)=\underbrace{n+n+\cdots +n}_{\text{$k$ copies}}-(1+2+\cdots k)=nk-\frac{k}{2}(k+1)$$

edited Nov 08 '13 at 16:32

Nick Peterson

32,430

answered Nov 08 '13 at 12:12

Shobhit

6,902

1

I edited your post to put the "underbrace" there; I think it makes this sort of thing more readable. If you don't like it, I won't be at all offended if you revert! – Nick Peterson Nov 08 '13 at 16:33
@NicholasR.Peterson Thanks for the edit. – Shobhit Nov 09 '13 at 02:11
In most cases that I encounter this, it's comparing all pairwise combinations in a set. EG A-Z (AB,AC, AD... AZ, BC, BD...) in this case, is k equal to n? – Adam Hughes Jul 18 '20 at 18:48

score 4 · Answer 2 · answered Nov 08 '13 at 12:12

4

Hint:

Try writing: $$ \sum_{k=1}^{n-1}k=\sum_{k=1}^{n-k-1}k+\sum_{k=n-k}^{n-1}k. $$ Your formula allows you to find the first two sums; subtraction should do the rest!

answered Nov 08 '13 at 12:12

Nick Peterson

32,430

score 3 · Answer 3 · answered Nov 08 '13 at 12:12

3

$ n k $ minus the sum from $1$ to $k$ equals

$$ n k - \frac12 k (k+1)$$

answered Nov 08 '13 at 12:12

Ron Gordon

138,521

What is the sum of (n-1)+(n-2)+...+(n-k)?

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