^{2}+ pq

^{3}+ .. + pq

^{i}+.. + pq

^{(n-1)}+ q

^{n}== 1

^{2}+ q

^{3}+ .. + q

^{i}+.. + q

^{(n-1)})+ q

^{n}== 1

^{n}== 1

^{n}will cancel from both formulae.

^{2}+ q

^{3}+ .. + q

^{i}+.. + q

^{(n-1)}) + q

^{n}== 1

^{n}+np

^{(n-1)}q ..

^{n}C

_{i}p

^{i}q

^{(n-i)}.. + npq

^{(n-1)}+ q

^{n}== 1

^{n}cancels, and we can divide through by p to get

^{2}+ q

^{3}+ .. + q

^{i}+.. + q

^{(n-1)}==

^{n-1}+np

^{(n-2)}q ..

^{n}C

_{i}p

^{(i-1)}q

^{(n-i)}.. + nq

^{(n-1)}

^{n})/(1-q) or (1-q

^{n})/p (as p+q=1). The formulae is thus

(1-q^{n})/p = sum(i=0..n-1) ^{n}C_{i} 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)} + p^{2}(p+q)^{(n-2)} + .. p^{i}(p+q)^{(n-i)} + .. p^{n}

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} = k^{n}

and thus

sum(i=0..n) ^{n}C_{i} p^{i}q^{(n-i)} = k^{n}

rearranging

sum(i=0..n) ^{n}C_{i} p^{(i-1)}q^{(n-i)} = (k^{n}-q^{n})/p

and so we can transform the above into

(p+q)^{n-1} + p(p+q)^{(n-2)} + p^{2}(p+q)^{(n-3)} + .. p^{i}(p+q)^{(n-i-1)} + .. p^{n-1} = (k^{n}-p^{n})/(k-p)

finally

k^{(n-1)} + pk^{(n-2)} + p^{2}k^{(n-3)} + .. p^{(n-2)}k + p^{(n-1)} = (k^{n}-p^{n})/(k-p)

An alternate method..

^{n}+ a

^{(n-1)}b + a

^{(n-2)}b

^{2}+ .. a

^{(n-i)}b

^{i}.. + b

^{n}

^{n+1}+ a

^{n}b + a

^{(n-1)}b

^{2}+ .. a

^{(n-i+1)}b

^{i}.. + ab

^{n}

^{n}b+ a

^{(n-1)}b

^{2}+ a

^{(n-2)}b

^{3}+ .. a

^{(n-i)}b

^{i+1}.. + b

^{n+1}

^{n+1}- b

^{n+1}

and so

^{n+1}- b

^{n+1})/(a-b)

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