find $n$ if $n=\sum _{k=1}^nk$

Question

anonymous · Answer

n=1 works, doesn't it?

$$\sum_{k=1}^{1}k = 1$$

anonymous · Answer

yes it does ,is that the only case

kirbykirby · Answer

I would think so. Can you imagine a sum 1+2+...+n = n, if n is other than 1?

anonymous · Answer

yeah...1+2 > 2     1 + 2 + 3 > 3 .....the sum is always gonna be bigger than n for the rest of the integers

ganeshie8 · Answer

n = n(n+1)/2

anonymous · Answer

the question further asks us to generalise $pk=\sum_{k=1}^nk$

anonymous · Answer

$p$  is a positive  integer

anonymous · Answer

i meant $$nk=\sum_{k=1}^nk$$according to @ganeshie8  set up
$pn=\dfrac{n(n+1)}{2}$
$2pn=n^2+n\implies n^2+n(1-2p)$
$$n=2p-1$$
so this holds for odd numbers

ganeshie8 · Answer

u meant this right :

$np=\sum \limits_{k=1}^nk$

anonymous · Answer

yes yes thank you @ganeshie8 @kirbykirby @BangkokGarrett

ganeshie8 · Answer

oh nice :) 
looking at  $n  =  n(n+1)/2$,   

n = 0 doesnt work since the lower bound is 1 is it ?

ganeshie8 · Answer

for ur first part of q..

anonymous · Answer

hmmmm yes ,actually it wpuld work for $np$ case as well

anonymous · Answer

i mean it doesnt work for both cases

ganeshie8 · Answer

Got it !! ty :)