Mathematical Series Problems

Try to solve as many of these math problems if you can. Good Luck!

1) What is the value of (1/2) * (2/3) * (3/4) * (4/5) * (5/6) * ... *(197/198) * (198/199) * (199/200)?

2) Compute 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(999*1000).

3) Find the value of (1/(sqrt(1) + sqrt(2)) + (1/(sqrt(2) + sqrt(3)) + (1/(sqrt(3) + sqrt(4)) + ...

+ (1/(sqrt(399) + sqrt(400)).

4) Let there exist a number x such that

x^0 + x^1 + x^2 + x^3 + x^4 + ... = 2020. If x can be expressed in the form p/q where p and q are relatively prime, what is p + q?

5) Compute (1/2^1) + (2/2^2) + (3/2^3) + (4/2^4) + ...

The answer to #1 is 1/200.

Correct! This one is pretty basic. You probably saw that the fractions cancel out when you multiply them. Great Job @SciFiChess !!!

The answer to #2 is 999/1000. I derived this formula myself without looking it up. Let n be a positive integer.

The answer to #5 is 2.

[1/(2¹)]+[2/(2²)]+[3/(2³)]+[4/(2⁴)]+…=

{[1/(2¹)]+[1/(2²)]+[1/(2³)]+[1/(2⁴)]+…}+

{[1/(2²)]+[1/(2³)]+[1/(2⁴)]+[1/(2⁵)]+…}+

{[1/(2³)]+[1/(2⁴)]+[1/(2⁵)]+[1/(2⁶)]+…}+

{[1/(2⁴)]+[1/(2⁵)]+[1/(2⁶)]+[1/(2⁷)]+…}+…=

1+(1/2)+(1/4)+(1/8)+…=2

The answer to #3 is 19. Let n be a positive integer and use the following formulas.

I solved #1, #2, #3 and #5 without looking at any formulas or other information.

I did look up this formula for #4. If x is a real number with -1<x<1, then

1+x+x²+x³+x⁴+…=1/(1-x)

In problem #4:

1/(1-x)=2020

x=2019/2020

p=2019

q=2020

p+q=4039

These are some Wikipedia links about mathematical series.

Series

Geometric series

Good Job @SciFiChess! The only thing is that I want you to explain a little more.

For example, for your formula for #4, can you please explain why it works? 

And for #7, please explain why for any positive number x where 0 < x < 1, that x^0 + x^1 + x^2 + ... then the result would be 1/(1-x)?

The answer to #1 is 1/200.

(1/2)(2/3)(3/4)…(197/198)(198/199)(199/200)

=(199!)/(200!)

=(199!)/[(199!)200]

=1/200

Problem 1 is already really easy to explain, and I pretty much stated the whole concept at #3.

I showed my work for problems 1 and 5, and I will give more details for the other problems as soon as I have time.

The answer to #3 is 19. Let n be a positive integer and use the following formulas.

Good Job!! You saw that you can rationalize the denominators of each term and then you would be left with a simple telescoping series. 

You can see more about telescoping math series in the following Wikipedia page:

Telescoping Series

Here is a Wikipedia page about rationalizing denominators:

Rationalisation