An algebra problem by U.N. Owen

$\begin{aligned} \sum_{j=1}^n (1+i)^j & = \frac{(1+i)[(1+i)^n-1]}{(1+i)-1} \\ & = \frac{(1+i)[(1+i)^n-1]}{i} \\ & = -i(1+i)[(1+i)^n-1] \\ & = (1-i)[(1+i)^n-1] \\ & = (1-i)(1+i)^n-1+i \\ & = (1-i^2)(1+i)^{n-1}-1+i \\ & = 2(1+i)^{n-1}-1+i \end{aligned}$

$\begin{aligned} \Rightarrow 31 + i & = 2(1 +i)^{n-1}-1+i \\ 2(1+i)^{n-1}-1 & = 31 \\ 2(1+i)^{n-1} & = 32 \\ (1+i)^{n-1} & = 16 \quad \quad \small \color{#3D99F6}{\text{Since } (1+i)^2 = 2i} \\ \Rightarrow (2i)^{\frac{n-1}{2}} & = 2^4 \\ 2^{n-1}i^{n-1} & = 2^8 \\ 2^8i^8 & = 2^8 \quad \quad \small \color{#3D99F6}{i^8=1} \\ \Rightarrow n-1 & = 8 \\ n & = 9 \space \boxed{\text{-- a multiple of 9}} \end{aligned}$

Use the formula for geometric progression that $S_n = \frac{a_{1}(r^{n}-1)}{r-1}$ to which in this question, $a_{1}=r=1+i$ . Plugging in the value gives $S_n = \frac{(1+i)((1+i)^{n}-1)}{(1+i)-1}$ or more simply $(1+i)^{n+1}+(i-1) = 31+i$ , which may be simplified to $(1+i)^{n+1} = 32$

From here we can use the properties of the absolute value, and by plugging in gives $|(1+i)^{n+1}| = |32|$ , or more simply $(\sqrt{2})^{n+1} = 32$ . Solve for $n$ gives $n = 9$