Solve for i in terms of other variables: F=C[frac((1+i)^n-1)(i)]

OlmekinjP 2021-09-15 Answered
Solve for i in terms of other variables:
\(\displaystyle{F}={C}{\left[{\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}\right]}\)

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Expert Answer

Alix Ortiz
Answered 2021-09-16 Author has 9249 answers

Step 1
since, we know Binomial Expansion for \(\displaystyle{\left({1}+{x}^{{n}}\right)}\) is Replace
\(\displaystyle{\left({1}+{x}\right)}^{{n}}={1}+{n}{x}+{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}!}}}{x}^{{2}}+{\frac{{{n}{\left({n}-{1}\right)}{\left({n}-{2}\right)}}}{{{3}!}}}{x}^{{3}}+\dot{{s}}+{x}^{{n}}\)
Replace n by (n-1), we get
\(\displaystyle{\left({1}+{x}\right)}^{{{n}-{1}}}={1}+{\left({n}-{1}\right)}{x}+{\frac{{{\left({n}-{1}\right)}{\left({n}-{2}\right)}}}{{{2}!}}}{x}^{{2}}+{\frac{{{\left({n}-{1}\right)}{\left({n}-{2}\right)}{\left({n}-{3}\right)}}}{{{3}!}}}{x}^{{3}}+\dot{{s}}+{x}^{{{n}-{1}}}\)
Step 2
Now, given that
\(\displaystyle{F}={C}{\left[{\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}\right]}\)
Since, we know
\(\displaystyle{\left({1}+{i}\right)}^{{n}}={1}+{n}{i}+{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}!}}}{i}^{{2}}+\dot{{s}}+{i}^{{n}}\)
then
\(\displaystyle{\left({1}+{i}\right)}^{{n}}-{1}={1}+{n}{i}+{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}!}}}{i}^{{2}}+\dot{{s}}+{i}^{{n}}-{1}\)
Now, we can cancel "1" in right side of equation, we get
\(\displaystyle{\left({1}+{i}\right)}^{{n}}-{1}={n}{i}+{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}!}}}{i}^{{2}}+\dot{{s}}\)
\(\displaystyle{\left({1}+{i}\right)}^{{n}}-{1}={i}{\left({n}{i}+{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}!}}}{i}^{{2}}+\dot{{s}}\right)}\) then
\(\displaystyle{\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}={\frac{{{i}{\left({n}{i}+{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}!}}}{i}^{{2}}+\dot{{s}}\right)}}}{{{i}}}}={\left({n}+{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}!}}}{i}+{\frac{{{n}{\left({n}-{1}\right)}{\left({n}-{2}\right)}}}{{{3}!}}}{i}^{{2}}+\dot{{s}}\right)}\)
\(\displaystyle{\left({\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}\right)}={\left({n}+{\frac{{{n}{\left({n}-{1}\right)}}}{{{2}!}}}{i}+{\frac{{{n}{\left({n}-{1}\right)}{\left({n}-{2}\right)}}}{{{3}!}}}{i}^{{2}}+\dot{{s}}\right)}\)
we can take common "n" in right side of equation, we get
\(\displaystyle{\left({\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}\right)}={n}{\left({1}+{\frac{{{\left({n}-{1}\right)}}}{{{2}!}}}{i}+{\frac{{{\left({n}-{1}\right)}{\left({n}-{2}\right)}}}{{{3}!}}}{i}^{{2}}+\dot{{s}}\right)}\)
\(\displaystyle{\left({\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}\right)}={n}{\left({1}+{\frac{{{1}}}{{{2}}}}{\left({\left({n}-{1}\right)}{i}+{\frac{{{\left({n}-{1}\right)}{\left({n}-{2}\right)}}}{{{2}!}}}{i}^{{2}}+\dot{{s}}+{1}-{1}\right)}\right)}\)
\(\displaystyle{\left({\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}\right)}={n}{\left({1}+{\frac{{{1}}}{{{2}}}}{\left({1}+{\left({n}-{1}\right)}{i}+{\frac{{{\left({n}-{1}\right)}{\left({n}-{2}\right)}}}{{{2}!}}}{i}^{{2}}+\dot{{s}}\right)}-{\frac{{{1}}}{{{2}}}}\right)}\)
From step (1), we know
\(\left(1+(n-1)i+\frac{(n-1)(n-2)}{2!}i^2+\dots\right)=(1+i) ^{n-1}\)\(\displaystyle{\left({\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}\right)}={n}{\left({\frac{{{1}}}{{{2}}}}+{\frac{{{\left({1}+{i}\right)}^{{{n}-{1}}}}}{{{2}}}}\right)}\)
then
\(\displaystyle{F}={C}{\left({\frac{{{\left({1}+{i}\right)}^{{n}}-{1}}}{{{i}}}}\right)}={n}{C}{\left({\frac{{{1}}}{{{2}}}}+{\frac{{{\left({1}+{i}\right)}^{{{n}-{1}}}}}{{{2}}}}\right)}\)
\(\displaystyle\Rightarrow{\frac{{{F}}}{{{n}{C}}}}-{\frac{{{1}}}{{{2}}}}={\frac{{{\left({1}+{i}\right)}^{{{n}-{1}}}}}{{{2}}}}\)
\(\displaystyle\Rightarrow{2}{\left({\frac{{{F}}}{{{n}{C}}}}-{\frac{{{1}}}{{{2}}}}\right)}={\left({1}+{i}\right)}^{{{n}-{1}}}\)
\(\displaystyle\Rightarrow{\left({\frac{{{2}{F}}}{{{n}{C}}}}-{1}\right)}^{{{\frac{{{1}}}{{{\left({n}-{1}\right)}}}}}}={1}+{i}\)
\(\displaystyle\Rightarrow{i}={\left({\frac{{{2}{F}}}{{{n}{C}}}}-{1}\right)}^{{{\frac{{{1}}}{{{\left({n}-{1}\right)}}}}}}-{1}\)
Answer:
\(\displaystyle{i}={\left({\frac{{{2}{F}}}{{{n}{C}}}}-{1}\right)}^{{{\frac{{{1}}}{{{\left({n}-{1}\right)}}}}}}-{1}\)

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