2
$\begingroup$

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?

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

$\endgroup$
10
  • 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
2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Still having issues clearing i step by step. $\endgroup$ Nov 27 '19 at 16:39
1
$\begingroup$

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

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

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

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

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.