From knowing $$7c=\sum_{i-1}^{50-c}k_i$$ and $c\choose 2 $=$ \sum_{i=1}^{50-c}$$k_i\choose 2 $ how can I get to $$\sum_{i=1}^{50-c}(k_i-\mu)^2=(50-c)\mu^2-14c\mu+c^2+6c$$ for some arbitrary $\mu$?

Question

From knowing $$7c=\sum_{i-1}^{50-c}k_i$$ and $c\choose 2 $=$  \sum_{i=1}^{50-c}$$k_i\choose 2 $ how can I get to $$\sum_{i=1}^{50-c}(k_i-\mu)^2=(50-c)\mu^2-14c\mu+c^2+6c$$ for some arbitrary $\mu$?

zzr0ck3r · Answer

The book I am reading has no explination, it just says 

"from the last two equations, it follows that for any real number $\mu$..." and then gives that last equality.

hartnn · Answer

you just need 
$\sum k_i^2 $
to be = c^2 +6c, right?
the rest is trivial

zzr0ck3r · Answer

it is late, I dont see the trivial part lol. But I have been staring at this for hours....

hartnn · Answer

$\sum \dfrac{k_i(k_i-1)}{2} = \dfrac{c(c-1)}{2}$

hartnn · Answer

ohh lol
$\sum k_i^2-2\sum k_i\mu +\sum \mu^2 $

zzr0ck3r · Answer

I didn't even remember we can do that, but sure it makes sense....

hartnn · Answer

$\mu $ is a constant so,
$\sum k_i^2-2\mu \sum k_i +\sum \mu^2 =\sum k_i^2-2\mu(7c)+\mu^2(50-c) $

hartnn · Answer

so coming back to $\sum k_i^2$
$\sum \dfrac{k_i(k_i-1)}{2} = \dfrac{c(c-1)}{2}$

$\sum  k_i^2 -\sum k_i = c^2 -c \ \sum k_i^2 = c^2-c+7c = ...$

just some algebraic manipulations:)

zzr0ck3r · Answer

wow

zzr0ck3r · Answer

you are the man

zzr0ck3r · Answer

I feel silly, but I'll take it :)

zzr0ck3r · Answer

this is for a 600 lvl graph theory class, and I can't to the algebra from precalculus ...

hartnn · Answer

its late, you'd surely have got it in the morning :P

rsadhvika · Answer

is there a way to work this combinatorially ?

zzr0ck3r · Answer

yeah, let's go with that.

zzr0ck3r · Answer

or right when I go to sleep....

zzr0ck3r · Answer

I have done enough counting for the night @rsadhvika  but dig in :)

rsadhvika · Answer

my browser is not diplaying latex

rsadhvika · Answer

$\sum\limits_{i=1}^n \dbinom{k_i}{2}$ is the total number of ways of choosing $2n$ people from $n$ different groups, picking $2$ people from each group.

rsadhvika · Answer

$\sum\limits_{i=1}^{n}k_i$ is the total number of people in all the groups.

rsadhvika · Answer

im using $50-c=n$ for simplicity..

rsadhvika · Answer

$\mu$  maybe number of ppl that you don't want to pick who have bad record maybe..

rsadhvika · Answer

but $\mu$ is fixed across all the groups

rsadhvika · Answer

that means you don't want to pick $\mu$ number of people from each group

rsadhvika · Answer

or maybe you're simply calculating RMS value of the observations and this may not have any useful meaning combinatorially..