# Clearing i on an equation

I'm having an issue clearing i on this equation, I've tried online step by step problem solvers but for some reason they give false as if there is no solution

This is how i write the equation on those sites to clear "i", any suggestion?

# $$98000=2350 \times \frac{(1+i)^{40}-1}{(1+i)^{40}i}$$

• Welcome to Mathematics Stack Exchange. Here's a hint: what is $(i+1)^2$? Nov 27 '19 at 16:15
• @J.W.Tanner isn't it 2i ? Nov 27 '19 at 16:18
• Yes; what's $(i+1)^4=((i+1)^2)^2$? Nov 27 '19 at 16:19
• @J.W.Tanner 4i = 4i, still missing the big picture though Nov 27 '19 at 16:20
• $(2i)^2=-4$; you should then be able to compute $(1+i)^{40}=((1+i)^4)^{10}$ fairly easily Nov 27 '19 at 16:21

I would suggest you compute $$(1+i)^2$$, and then you should be able to compute $$(1+i)^{40}$$ fairly easily.

• Still having issues clearing i step by step. Nov 27 '19 at 16:39

$$(1+i)^2 = 1+2i-1 = 2i$$

$$(1+i)^4 = ((1+i)^2)^2 =(2i)^2=-4$$

$$(1+i)^{40}=((1+i)^{4})^{10}=(-4)^{10}=2^{20}$$

$$\frac{(1+i)^{40}-1}{(1+i)^{40}i}=\frac{1}{i}-\frac{1}{(1+i)^{40}i}$$

Substituting $$(1+i)^{40}=2^{20}$$ and $$1/i=-i$$

$$\implies-i+i(2^{-40})$$

$$\implies i(2^{-40}-1)$$

This is a pure imaginary number. Your equation in the question is wrong.

• So there's no solution, i was under the idea the result should be 0,01524, but i find no way in which it can give that result. :/ Nov 27 '19 at 17:05
• More like false statement. Solution is when you have a variable in the equation.
– xax
Nov 27 '19 at 17:07