```Question 262794
The original problem is
(i) {{{r=(r-p*i)(1+i)^n}}}
where i I assume is a standard variable and not an imaginary number.
Notice that (1+i)^n is a binomial that can be expanded as
(nC0)*1^n*i^0 +(nC1)*1^(n-1)*i^1  + (nC2)*1^(n-2)*i^2 + (nC3)*1^(n-3)*i^3 + . . . +
(nC(n-1))*1^1*i^(n-1) + (nCn)*1^0*i^n
This makes the problem a bit more interesting.
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If n = 1,
then
(ii) {{{r = (r-p*i)(1+i)^1}}}
and expanded, we get
(iii) {{{r = r + r*i - p*i - p*i^2}}}
setting = 0 and combining like terms we get
(iv) {{{ 0 = p*i^2 + (p-r)i + 0}}}
using the quadratic, we get
i = (r-p)/p
or
i = 0. This is considered a trivial solution in that it works for all values of n.
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If n = 2,
then
(v) {{{r = (r-p*i)(1+i)^2}}}
and expanded, we get
(vi) {{{r = r + 2r*i + 2i^2 - p*i - 2p*i^2 - p*i^3}}}
setting = 0 and combining like terms we get
(vii) {{{ 0 = p*i^3 + (2p-r)i^2 + (p-2r)i}}}
factoring out the i, we get
(viii) {{{ 0 = i(p*i^2 + (2p-r)i^1 + (p-2r))}}}
using the quadratic, we get
i = ((r-2p) +-sqrt(r^2+4pr))/2p
or
i = 0. This is considered a trivial solution in that it works for all values of n.
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We are looking for a pattern here. We know that i = 0 must always be answer.
The other values of i appear to change.
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What will happen is that (v) and (viii) will have
1 trivial value at i = 0.
1 non trivial value that follows binomial expansion:
(pi^n + ((nc1)p - (ncn)r)i^(n-1) + ((nc2)p - (nc(n-1)r)i^(n-2) + ((nc3)p - (nc(n-2)r)i^(n-3) +
. . . + ((nc(n-1))p - (nc2)r)i^(1) + ((nc(n))p - (nc1)r)i^(0))
which ccan be difficult to factor not knowing p and r.
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I hope this helps . . . ```