Question

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

Answer 1

n=1 works, doesn't it?

\[\sum_{k=1}^{1}k = 1\]

Answer 2

yes it does ,is that the only case

Answer 3

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

Answer 4

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

Answer 5

n = n(n+1)/2

Answer 6

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

Answer 7

\(p\) is a positive integer

Answer 8

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

Answer 9

u meant this right :

\(np=\sum \limits_{k=1}^nk\)

Answer 10

yes yes thank you @ganeshie8 @kirbykirby @BangkokGarrett

Answer 11

oh nice :)
looking at \(n = n(n+1)/2\),

n = 0 doesnt work since the lower bound is 1 is it ?

Answer 12

for ur first part of q..

Answer 13

hmmmm yes ,actually it wpuld work for \(np\) case as well

Answer 14

i mean it doesnt work for both cases

Answer 15

Got it !! ty :)