#S= 1/1×2 + 1/3×2 + 1/3×4 +⋯+ 1/99×100
opal hollowBOT
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neat wing
what is your question ? To me, S = Σ_{n=1}^99 1/(n(n+1)), and your argument works
daring isle
i don't understand how ( 1/1 − 1/2)+( 1/3-1/4)+...+(1/99-1/100) can be equal to 99\100 like what steps i need to do to turn this (1/1 − 1/2)+( 1/3-1/4)+...+(1/99-1/100) to the answer, i know that some how most of this numbers will be canceled but idk how.
neat wing
reread the definition of S
S = 1/(1×2) + 1/(2×3) + 1/(3×4) + … + 1/(99×100)
they just sneakily wrote 1/(3×2) instead of 1/(2×3)
@daring isle
crude bear
i don't understand how ( 1/1 − 1/2)+( 1/3-1/4)+...+(1/99-1/100) can be equal to ...
these are called telescoping series, and this particular one can be evaluated using partial fractions
we have $\sum_{n=1}^{99}\frac{1}{n(n+1)}$
fiery mossBOT
mark datter
crude bear
what we want to do is make it telescoping: i.e., make the summand more simple so we can cancel out the terms
you can do this by splitting the summand into partial fractions
so, turn $\frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$, then you’ll notice something nice about the sum if you rewrite it in this form
fiery mossBOT
mark datter
neat wing
@crude bear if you read what they wrote they already noticed that…
they are just confused about the 1/(1×2) + 1/(3×2) + …
dusky cave
**mark datter**
wait this looks like the sum of n natural numbers
crude bear
they are just confused about the 1/(1×2) + 1/(3×2) + …
i dont necessarily agree but i do admit i misread
i think they're just confused about why the numbers cancel the way they do
novel tinsel
Notice that 1/n - 1/(n+1) = 1/(n(n+1))
crude bear
Notice that 1/n - 1/(n+1) = 1/(n(n+1))
they already covered that as petaire said
they just dont understand the concept
reef kelpBOT
@daring isle