Pushy

So, when translating to LaTeX, we get \(x = 1 + \frac{x}{1 + \frac{x}{1+ \frac{x}{1 + \cdots}}}\)

Look at all the fractions under the first x. That is \(1 + \frac{x}{1 + \frac{x}{1+ \frac{x}{1 + \cdots}}}\)

But that is also x! 

So, x = 1 + x/x which is x = 1 + 1 which is 2.

Thanks :)

Thanks :)

Sorry for all these questions XD Plz forgive me XD. But thanks :)

Thanks :)

Thanks :)

So, we know that a^2 + b^2 + c^2 = 13.

Also, ab + bc + ca = 6.

And we want to find a + b + c. So, we sqaure a + b + c.

That would be a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2.

We group the squares together to get a^2 + b^2 + c^2, which is 13.

Then we group the ab or bc or ac together.

That gets us 2(ab + bc + ac) which is 2*6 which is 12.

Then we add to get 13 + 12 = 25.

Thanks :)

So, we can plug the variables into the equation. 

\(1- i = (1+3i) I.\)

We divide 1 + 3i on both sides to get:

\(\frac{1-i}{1+3i} = I\)

Then we multiply 1-3i on the numerator and the denominator of the left side to get:

\(-\frac12-i\).

So, \(I = -\frac12-i\)

So 

i^1 = i,

i ^ 2 = -1,

i^3 = -i,

i^4= 1,

i ^ 5 = i,

and it keeps repeating between the four (i,-1,-i,i).

So, to find i^11, we find 11mod4 (or the remainder when dividing 11 and 4) to get 3. So, i ^11 is -i.

Then you do the rest.

i^16 = 1

i^21 = i

i^26 = -1

i^31 = -i.

So, all of those added together is -i + 1 + i - 1 - i which is -i.

Wow... that's a great solution! Thanks :)

It's a giant calculator which you can ask questions on. Like you did here. People will answer to them and... yeah :)

Wow that's complicated... but thanks for helping me and taking the time to do it :)

I don't want to sound rude, but like... people come here because they need help. People who post answers have an explanation so they can learn. :)