im completely lost in my advanced calc class. i would really appreciate any help

im sorry in advance for my lack of proper notation (i have no idea how do write this on my cpu)

i took and did well at calc I and II last year. but i got accepted into an advanced calc program at my college. were relearning everything from the beginning but this time there we cant go straight into doing things, we have to proove everything---for example, if it says to integrate, were expected to do it step by step so 'an alien' can understand it, as opposed to the normal shorter way..anyways, hopefully someone knows what im talking about

ive worked for the past couple hours on these and got just over half. heres the half that im completely clueless on

1. derive this formula using basic properties of summation notation if possible (additive, homogenous, telescoping are the 3 possibilities)

summation from k=0 to n ... x^k = (1-x^(n+1))/(1-x) if x doesnt equal 1

[hint ... (1-x) * summation notation from k=0 to n ... x^k]
b. what is the sum equal to when x=1

hopefully you can decipher what im trying to transcribe-the rest arent that hard to understand

2. prove the following of absolute values
abs [ abs(x) - abs (y) ] </= abs (x - y)

3.compute the value of the following integrals. the notation [x] denotes the greatest integer </= x
(im trying to write less than or equal to)
a. integral from 3 to -1 2[x] dx
b. integral from 3 to -1 [2x] dx
c. integral from 3 to -1 [-x] dx

5. give an example of a step function s defined on [0,5]which has the following properties...integral from 2 to 0 s(x) dx= 5
integral from 5 to 0 s(x) dx= 2

6. if f(x)= integral from x to 0 [t] dt for x greater than or equal to 0, draw the graph of f over the interval [0,4]

i got some answers if it helps but i need to figure out how to get them
1b. n+1
3a.4
3b. 6
3c. -6

i think im in over my head here, so any help, answers, or explanations is very appreciated

For number 1:
$ \sum_{k=0}^n x^k $

If you multiply with (1-x), assuming $x \neq 1$ you can turn it into a telescoping series.

$ (1-x)\sum_{k=0}^n x^k = \sum_{k=0}^n x^k - \sum_{k=0}^n x^{k+1} = 1 - x^{k+1} $
Then you divide with (1-x) on both sides and get:

$ \sum_{k=0}^n x^k =\frac{1 - x^{k+1}}{1-x} $
Is this one clear?

for number 2: we use the triangle inequality:

$ |x| = |x - y + y| \leq |x-y| + |y| $
$ |x| - |y| \leq |x-y| \\ $

$ |y| = |y-x+x| \leq |y-x| + |x| $
$ |y|-|x| \leq |y-x| = |x-y| $

Now since $|y|-|x| \leq |x-y| \text{ and } |x| - |y| \leq |x-y|$ it proves that
$ ||x| - |y|| \leq |x-y| $

Do you understand ?

thanks alot!!!

some quick ?s

on the bottom of number 2, how can you say x-y= y- x

I did not say that x-y= y- x, but rather |x-y|= |y- x|. It follows directly from defenition.