(1-qn)/p = sum(i=0..n-1) nCi p(i-1)q(n-i)
Looking carefully at the above this is merely a re-write of (p+q)n .
This can also be broken down to look like the following..
(p+q)n + p(p+q)(n+1) + p2(p+q)(n-2) + .. pi(p+q)(n-i) + .. pn
So we are now back to our original formulae as p+q=1!!
However, lest assume p+q!=1. let us assume p+q=K instead, and use a similar method (that I stumbled upon by chance) to see where it leads. so
(p+q)n = kn
sum(i=0..n) nCi piq(n-i) = kn
sum(i=0..n) nCi p(i-1)q(n-i) = (kn-qn)/p
and so we can transform the above into
(p+q)n-1 + p(p+q)(n-2) + p2(p+q)(n-3) + .. pi(p+q)(n-i-1) + .. pn-1 = (kn-pn)/(k-p)
k(n-1) + pk(n-2) + p2k(n-3) + .. p(n-2)k + p(n-1) = (kn-pn)/(k-p)
An alternate method..