10 grade math

[radiant quail]

i got no idea how to start this problem at all

[cyan crest]

try looking at each term on its own, see if you can see a pattern for:

$\cfrac{\frac{1}{2}}{1+\frac{1}{2}}$

$\cfrac{\frac{1}{3}}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)}$

$\cfrac{\frac{1}{4}}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)}$

etc...

Once you find the pattern, can you express each term as a difference?

I solved it using a telescoping series, FYI.

[pure linden]

Jay

[lost path]

Even if you don't know about telescoping sums, if you find the first 3 or 4 partial sums you can probably identify the pattern.

[radiant quail]

hey so i do got the pattern but my problem with this question is that i dont really know what to do, sure i simplyfied it to 1/3 + 1/6 + 1/10 + 1/15 but like how can i even simplyfired it to a/b then do a + b

oh wiat

i think i might got it

well no i didnt

damn

alright i got

1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/81

then i simplyfied it to

1/2 + 1/6 + 1/12 + 1/20

i think i just kept going until 2023

but how would i know when to end?

man this is a pain

@cyan crest hey man can i see how you did it?

[lost path]

Start adding them up and look for a pattern

[cyan crest]

Look at rearranging the biggest term along the way used on the smaller ones...

$\cfrac{\frac{1}{2023}}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\times ... \times \left(1+\frac{1}{2023}\right)}$

$=\cfrac{\frac{1}{2023}}{\frac{2+1}{2}\times\frac{3+1}{3}\times\frac{4+1}{4}\times ... \times \frac{2022+1}{2022}\times\frac{2023+1}{2023}}$

$=\cfrac{\frac{1}{2023}}{\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times ... \times \frac{2023}{2022}\times\frac{2024}{2023}}$

$=\cfrac{\frac{1}{2023}}{\frac{1}{2}\times\frac{2024}{1}}$ after cancelling

$=\frac{1}{2023} \times \frac{2}{2024}$

[pure linden]

Jay

[cyan crest]

And all the terms follow the same pattern:

$\cfrac{\frac{1}{2}}{1+\frac{1}{2}} = \frac{2}{2 \times 3}$

$\cfrac{\frac{1}{3}}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)} = \frac{2}{3 \times 4}$

Thus, we are trying to find $\sum_{k=2}^{2023} \frac{2}{k(k+1)}$