Algebra Problem By Yahia El Haw

$\begin{aligned} S & = (1+i)+(1+i)^2+(1+i)^3 + ...+(1+i)^{100} \\ & = (1+i) \left(\frac {(1+i)^{100}-1}{1+i-1} \right) \\ & = (1+i) \left(\frac {(1+i)^{2 \times 50}-1}i \right) \\ & = i(1+i) \left(\frac {(2i)^{50}-1}{i^2} \right) \\ & = (i+i^2) \left(\frac {2^{50}i^{50}-1}{i^2} \right) \\ & = (i-1) \left(\frac {-2^{50}-1}{-1} \right) \\ & = (2^{50} + 1)(-1+i) \end{aligned}$

$\implies a + b + 1 + d + e = 2+50+1-1+1 = \boxed{53}$

Set $z = (1 + i)$ . Then, the sum can be expressed as $\begin{array}{rl} z + z^2 + \cdots + z^{100} &= \sum\limits_{n = 1}^{100} z^n\\ &= \left(\sum\limits_{n = 0}^{100} z^n\right) - 1\\ &= \dfrac{z^{101} - 1}{z - 1} - 1\\ &= \dfrac{z\left(z^{100} - 1\right)}{z - 1} \end{array}$ Converting $(1 + i)$ in polar coordinates, $\begin{array}{rl} z = 1 + i &= \sqrt{2} \cdot \dfrac{1}{\sqrt{2}} + \sqrt{2} \cdot \dfrac{i}{\sqrt{2}}i\\ &= \sqrt{2}\left(\dfrac{1}{\sqrt{2}} + \dfrac{i}{\sqrt{2}}\right)\\ &= \sqrt{2}\left(\cos\left(\dfrac{\pi}{4}\right) + i\sin\left(\dfrac{\pi}{4}\right)\right) \end{array}$ By DeMoivre's Theorem , $\begin{array}{rl} z^{100} &= \sqrt{2}^{100}\left(\cos\left(\dfrac{\pi}{4}\right) + i\sin\left(\dfrac{\pi}{4}\right)\right)^{100}\\ &= 2^{50}\left(\cos\left(\dfrac{100\pi}{4}\right) + i\sin\left(\dfrac{100\pi}{4}\right)\right)\\ &= 2^{50}\left(\cos\left(25\pi\right) + i\sin\left(25\pi\right)\right)\\ &= -2^{50} \end{array}$ Thus, $\begin{array}{rl} \dfrac{z}{z - 1}\left(z^{100} - 1\right) &= \dfrac{1 + i}{i}\left(-2^{50} - 1\right)\\ &= -\left(2^{50} + 1\right)\left(\dfrac{1 + i}{i} \cdot \dfrac{i}{i}\right)\\ &= -\left(2^{50} + 1\right)\left(1 - i\right)\\ &= \left(2^{50} + 1\right)\left(i - 1\right) \end{array}$ Comparing the above expression with the given form, we have $a + b + 1 + d + e + f = \boxed{53}$ .