0
$\begingroup$

I came across this problem that I couldn't really find an effective approach for. It was the simplification of $(i+1)^{2010} \text{ }- (i-1)^{2010}.$

I know that this roughly translates to $ (i^{2010} + i^{2019} + ... i + 1) \text{ }$ minus a similar expression, and I also know that the answer is $2^{1006}i$, but I just don't know any other steps that I could take.

$\endgroup$
1
$\begingroup$

$$(i\pm1)^2=\pm2i$$

$$(i\pm1)^{4n+2}=(\pm2i)^{2n+1}=\pm2i((\pm2i)^2)^n=\pm2i(-4)^n$$

$\endgroup$

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.