#### Monoxdifly

Determine the value of $$\displaystyle \frac13+\frac16+\frac1{10}+\frac1{15}+\frac1{21}+...+\frac1{231}$$
I know that it means $$\displaystyle \frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+\frac1{1+2+3+4+5}+\frac1{1+2+3+4+5+6}+...+\frac1{231}$$, but how do I answer? It's from a student worksheet for 7th graders, so they haven't learnt about sequence and series yet. Granted, the book also says that the question was from a middle school-level math contest.

#### romsek

MHF Helper
$\sum \limits_{k=2}^{21} ~\dfrac{1}{\sum \limits_{j=1}^k j} = \sum \limits_{k=2}^{21} \dfrac{2}{k(k+1)} = \\~\\ 2 \sum \limits_{k=2}^{21} \dfrac{1}{k} - \dfrac{1}{k+1} = \\~\\ 2 \left(\dfrac 1 2 - \dfrac 1 3 + \dfrac 1 3 - \dfrac 1 4 + \dots - \dfrac{1}{22}\right) = \\ 2\left(\dfrac 1 2 - \dfrac{1}{22}\right) = \dfrac{10}{11}$

topsquark

#### Monoxdifly

That confirms me that it's definitely not supposed to be for a 7th grader.

topsquark

#### romsek

MHF Helper
Nothing simpler is really jumping out at me.

#### topsquark

Forum Staff
A seventh grade Math problem (in a presumably American school) and you don't know how to do it in a simple manner? Where have you guys been? They use calculators.

-Dan

#### Debsta

MHF Helper
Interesting pattern when you look at the sums:

First 2: $$\displaystyle \frac{1}{3}+\frac{1}{6} =\frac{1}{2}$$

First 3: $$\displaystyle \frac{1}{3}+\frac{1}{6} +\frac{1}{10}=\frac{3}{5}$$

First 4: $$\displaystyle \frac{2}{3}$$

First 5: $$\displaystyle \frac{5}{7}$$

First 6: $$\displaystyle \frac{3}{4}$$

First 7: $$\displaystyle \frac{7}{9}$$

Note the two "patterns" for sums of evens and odds.
Since 231 is the 20th number in the triangular sequence (starting with 3 ie, 3, 6, 10, ...), we are particularly interested in the sum of evens.

First 2: $$\displaystyle \frac{1}{2}$$ ... numerator is half of 2, denominator is 1 more

First 4: $$\displaystyle \frac{2}{3}$$ ... numerator is half of 4, denominator is 1 more

First 6: $$\displaystyle \frac{3}{4}$$ ... numerator is half of 6, denominator is 1 more
.
.
.
First 20: $$\displaystyle \frac{10}{11}$$ ... numerator is half of 20, denominator is 1 more

Beautiful!!

#### Monoxdifly

Interesting pattern when you look at the sums:

First 2: $$\displaystyle \frac{1}{3}+\frac{1}{6} =\frac{1}{2}$$

First 3: $$\displaystyle \frac{1}{3}+\frac{1}{6} +\frac{1}{10}=\frac{3}{5}$$

First 4: $$\displaystyle \frac{2}{3}$$

First 5: $$\displaystyle \frac{5}{7}$$

First 6: $$\displaystyle \frac{3}{4}$$

First 7: $$\displaystyle \frac{7}{9}$$

Note the two "patterns" for sums of evens and odds.
Since 231 is the 20th number in the triangular sequence (starting with 3 ie, 3, 6, 10, ...), we are particularly interested in the sum of evens.

First 2: $$\displaystyle \frac{1}{2}$$ ... numerator is half of 2, denominator is 1 more

First 4: $$\displaystyle \frac{2}{3}$$ ... numerator is half of 4, denominator is 1 more

First 6: $$\displaystyle \frac{3}{4}$$ ... numerator is half of 6, denominator is 1 more
.
.
.
First 20: $$\displaystyle \frac{10}{11}$$ ... numerator is half of 20, denominator is 1 more

Beautiful!!
Ah, I see. I hope the one I am tutoring also finds this beautiful.