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?

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

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

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

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

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



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


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

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

  • $\begingroup$ 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. :/ $\endgroup$ Nov 27 '19 at 17:05
  • $\begingroup$ More like false statement. Solution is when you have a variable in the equation. $\endgroup$
    – xax
    Nov 27 '19 at 17:07

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