document.write( "Question 262794: I'm trying to solve for \"i\" in the expression:\r
\n" ); document.write( "\n" ); document.write( "r=(r-p*i)(1+i)^n \n" ); document.write( "
Algebra.Com's Answer #193569 by drk(1908)\"\" \"About 
You can put this solution on YOUR website!
The original problem is
\n" ); document.write( "(i) \"r=%28r-p%2Ai%29%281%2Bi%29%5En\"
\n" ); document.write( "where i I assume is a standard variable and not an imaginary number.
\n" ); document.write( "Notice that (1+i)^n is a binomial that can be expanded as
\n" ); document.write( "(nC0)*1^n*i^0 +(nC1)*1^(n-1)*i^1 + (nC2)*1^(n-2)*i^2 + (nC3)*1^(n-3)*i^3 + . . . +
\n" ); document.write( "(nC(n-1))*1^1*i^(n-1) + (nCn)*1^0*i^n
\n" ); document.write( "This makes the problem a bit more interesting.
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\n" ); document.write( "If n = 1,
\n" ); document.write( "then
\n" ); document.write( "(ii) \"r+=+%28r-p%2Ai%29%281%2Bi%29%5E1\"
\n" ); document.write( "and expanded, we get
\n" ); document.write( "(iii) \"r+=+r+%2B+r%2Ai+-+p%2Ai+-+p%2Ai%5E2\"
\n" ); document.write( "setting = 0 and combining like terms we get
\n" ); document.write( "(iv) \"+0+=+p%2Ai%5E2+%2B+%28p-r%29i+%2B+0\"
\n" ); document.write( "using the quadratic, we get
\n" ); document.write( "i = (r-p)/p
\n" ); document.write( "or
\n" ); document.write( "i = 0. This is considered a trivial solution in that it works for all values of n.
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\n" ); document.write( "If n = 2,
\n" ); document.write( "then
\n" ); document.write( "(v) \"r+=+%28r-p%2Ai%29%281%2Bi%29%5E2\"
\n" ); document.write( "and expanded, we get
\n" ); document.write( "(vi) \"r+=+r+%2B+2r%2Ai+%2B+2i%5E2+-+p%2Ai+-+2p%2Ai%5E2+-+p%2Ai%5E3\"
\n" ); document.write( "setting = 0 and combining like terms we get
\n" ); document.write( "(vii) \"+0+=+p%2Ai%5E3+%2B+%282p-r%29i%5E2+%2B+%28p-2r%29i\"
\n" ); document.write( "factoring out the i, we get
\n" ); document.write( "(viii) \"+0+=+i%28p%2Ai%5E2+%2B+%282p-r%29i%5E1+%2B+%28p-2r%29%29\"
\n" ); document.write( "using the quadratic, we get
\n" ); document.write( "i = ((r-2p) +-sqrt(r^2+4pr))/2p
\n" ); document.write( "or
\n" ); document.write( "i = 0. This is considered a trivial solution in that it works for all values of n.
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\n" ); document.write( "We are looking for a pattern here. We know that i = 0 must always be answer.
\n" ); document.write( "The other values of i appear to change.
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\n" ); document.write( "What will happen is that (v) and (viii) will have
\n" ); document.write( "1 trivial value at i = 0.
\n" ); document.write( "1 non trivial value that follows binomial expansion:
\n" ); document.write( "(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) +
\n" ); document.write( ". . . + ((nc(n-1))p - (nc2)r)i^(1) + ((nc(n))p - (nc1)r)i^(0))
\n" ); document.write( "which ccan be difficult to factor not knowing p and r.
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\n" ); document.write( "I hope this helps . . . \n" ); document.write( "
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