For what value of \(n\) is \(i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i\) ?
\(\begin{array}{|rcll|} \hline s &=& i + 2i^2 + 3i^3 + \cdots + ni^n \\ \hline \end{array}\)
\(\begin{array}{|rclll|} \hline s &=& i + 2i^2 + 3i^3 +4i^4+ \cdots + ni^n \\ is &=& \qquad i^2 +2i^3+3i^4+\ldots (n+1)i^n+ni^{n+1} \\ \hline s-is &=& \underbrace{ i+ i^2+i^3+i^4+\ldots +i^n}_{=S~ (GP)}-ni^{n+1}\\ s(1-i) &=& S-ni^{n+1}\\ &&\begin{array}{|rclll|} \hline S &=& i+ i^2+i^3+i^4+\ldots +i^n \\ iS &=& \qquad i^2+i^3+i^4+\ldots +i^n +i^{n+1} \\ \hline S - iS &=& i-i^{n+1} \\ S (1- i) &=& i( 1-i^{n}) \\ S &=& \dfrac{i( 1-i^{n})}{1-i} \\ \hline \end{array} \\\\ s(1-i) &=& \dfrac{i( 1-i^{n})}{1-i}-ni^{n+1} \\\\ s &=& \dfrac{i( 1-i^{n})}{(1-i)^2}-\dfrac{ni^{n+1}}{ 1-i } \\\\ s &=& \dfrac{i( 1-i^{n})}{(1-i)^2}-\left(\dfrac{ni^{n+1}}{ 1-i }\right)\cdot \left(\dfrac{1-i}{1-i}\right) \\\\ s &=& \dfrac{i( 1-i^{n}) -ni^{n+1}(1-i) }{(1-i)^2} \\\\ s &=& \dfrac{i( 1-i^{n}) -ni^{n}i(1-i) }{(1-i)^2} \\\\ s &=& \dfrac{i\Big( 1-i^{n} -ni^{n}(1-i)\Big) }{(1-i)^2} \quad | \quad (1-i)^2 = -2i \\\\ s &=& \dfrac{i\Big( 1-i^{n} -ni^{n}(1-i)\Big) }{-2i} \\\\ s &=& \dfrac{ 1-i^{n} -ni^{n}(1-i) }{-2 } \\\\ s &=& \dfrac{ -1+i^{n} +ni^{n}(1-i) }{2 } \\\\ s &=& \dfrac{ -1+i^{n} +ni^{n} -ni^{n+1} }{2 } \\\\ \mathbf{s} &=& \mathbf{\dfrac{ i^{n}(1+n) -i^{n+1}n -1 }{2 }} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{s} = \mathbf{\dfrac{ i^{n}(1+n) -i^{n+1}n -1 }{2 }} &=& 48 + 49i \\\\ i^{n}(1+n) -i^{n+1}n -1 &=& 96 + 98i \\\\ \mathbf{{\color{red}i^{n}}(1+n) -{\color{red}i^{n+1}}n } &=& \mathbf{97 + 98i} \\ \hline \end{array} \)
We compare the sides of the equation and note that i^n and i^(n+1) are adjacent values.
This means that one is either +i or -i and the other is either +1 or -1.
So there are two solutions for the comparison.
1. First possible solution
\(\begin{array}{|rcll|} \hline 98i &=&{ -\color{red}i^{n+1}}n \\ 97 &=& {\color{red}i^{n}}(1+n) \\ \hline \dfrac{98i}{97} &=& \dfrac{-i^{n+1} n}{i^{n} (1+n)} \\\\ \dfrac{98i}{97} &=& \dfrac{-i^{n}i\cdot n}{i^{n} (1+n)} \\\\ \dfrac{98i}{97} &=& \dfrac{- i\cdot n}{ (1+n)} \\\\ \dfrac{98}{97} &=& \dfrac{-n}{ (1+n)} \\\\ (1+n)98 &=& -97n \\ 98+98n &=& -97n \\ 195n &=& -98 \\ 195n &=& -98 \\ n &=& -\dfrac{98}{195} \\ n &=& -0.50256410256 \qquad \text{no solution, n is no integer } \\ \hline \end{array}\)
2. Second possible solution
\(\begin{array}{|rcll|} \hline 98i &=& {\color{red}i^{n}}(1+n) \\ 97 &=&{ -\color{red}i^{n+1}}n \\ \hline \dfrac{98i}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n+1} n} \\\\ \dfrac{98i}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n}i\cdot n} \\\\ \dfrac{98i^2}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n} \cdot n} \\\\ \dfrac{-98}{97} &=& \dfrac{1+n}{-n} \\\\ \dfrac{98}{97} &=& \dfrac{1+n}{n} \\\\ 98n &=& 97(1+n) \\ 98n &=& 97 + 97n \\ \mathbf{n} &=& \mathbf{97} \\ \hline \end{array}\)
\(i + 2i^2 + 3i^3 + \cdots + 97i^{97} = 48 + 49i\)
