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[ASK] Addition of Fractions: 1/3+1/6+1/10+1/15+1/21+...+1/231

Monoxdifly

Well-known member
Aug 6, 2015
271
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,682
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.
A 7th grader should start by asking how $\dfrac1{231}$ fits into the series of terms $\dfrac1{1+2+\ldots + n}$. A bright student like a middle school math contestant should probably know that $1+2+\ldots + n = \frac12n(n+1)$. If that is equal to $231$ then $n(n+1) = 462$, from which it's not hard to see that $n=21$.

So we want to find the sum \(\displaystyle \frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+ \ldots +\frac1{1+2+3+4+ \ldots +21}\).

Start by summing the first few terms, adding one more term to the previous sum each time: $$\frac1{1+2} = \frac13,$$ $$\frac1{1+2}+\frac1{1+2+3} = \frac13 + \frac16 = \frac36 = \frac12,$$ $$\frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4} = \frac12 + \frac1{10} = \frac6{10} = \frac35,$$ $$\frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+\frac1{1+2+3+4+5} = \frac35 + \frac1{15} = \frac{10}{15} = \frac23,$$ $$\frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+\frac1{1+2+3+4+5}+\frac1{1+2+3+4+5+6} = \frac23 + \frac1{21} = \frac{15}{21} = \frac57.$$ At this point, a bright 7th grader ought to spot a pattern in these partial sums $$\frac1{1+{\color{red}2}} = \color{red}\frac13,$$ $$\frac1{1+2}+\frac1{1+2+{\color{red}3}} = \frac12 = \color{red}\frac24,$$ $$\frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+{\color{red}4}} = \color{red}\frac35,$$ $$\frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+\frac1{1+2+3+4+{\color{red}5}} = \frac23 = \color{red}\frac46,$$ $$\frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+\frac1{1+2+3+4+5}+\frac1{1+2+3+4+5+{\color{red}6}} = \color{red}\frac57,$$ and jump to the (corrrect) conclusion that \(\displaystyle \frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+ \ldots +\frac1{1+2+3+4+ \ldots + {\color{red}21}} = {\color{red}\frac {20}{22}} = \frac{10}{11}\).

I think it would be asking too much of a 7th grader to do any more than that. But a user of this forum ought to be able to use induction to prove the result that \(\displaystyle \sum_{n=2}^N \frac1{1+2+\ldots+n} = \frac{N-1}{N+1}.\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,713
Indeed. A standard trick for high school math contests is to start with one term, then two terms, and so on, and try to spot a pattern. Then generalize the pattern, and fill in the final value.
And yes, it requires a bit of a jump to spot the pattern that Opalg pointed out.

Anyway, I'd like to point out a different method that should be within reach of high schoolers, which admittedly does become a bit easier with knowledge of and familiarity with fractions in general, arithmetic sequences, telescoping sums, and fraction decomposition.
If finds Opalg 's formula directly rather than spotting the pattern and proving it by induction.

Let $n$ be such that $1+2+\ldots+n = 231$, then:
$$
\begin{aligned}
\frac 13 + \frac 16 + \frac 1{10}+\ldots+ \frac 1{231}
&= \frac 1{1+2} + \frac 1{1+2+3} + \frac 1{1+2+3+4}+\ldots+ \frac 1{1+2+\ldots+n} \\
&= \frac 1{\frac 12\cdot 2(1+2)} + \frac 1{\frac 12\cdot 3(1+3)} + \frac 1{\frac 12\cdot 4(1+4)}+\ldots+ \frac 1{\frac 12\cdot n(1+n)} \\
&= 2\Bigg(\frac 1{2\cdot 3} + \frac 1{3\cdot 4} + \frac 1{4\cdot 5}+\ldots+ \frac 1{n(n+1)}\Bigg) \\
&= 2\Bigg(\left(\frac 12-\frac 13\right) + \left(\frac 13-\frac 14\right) + \left(\frac 14-\frac 15\right)+\ldots+ \left(\frac 1n-\frac 1{n+1}\right)\Bigg) \\
&= 2\Bigg(\frac 12-\frac 1{n+1}\Bigg) \\
&= \frac {n-1}{n+1}
\end{aligned}
$$
That leaves figuring out what $n$ is.
We can solve the equation, or we can just use trial and error to find that $n=21$. Indeed $\frac 12\cdot 21(1+21) = 231$.

So the sum is:
$$\frac 13 + \frac 16 + \frac 1{10}+\ldots+ \frac 1{231}=\frac {21-1}{21+1}=\frac{10}{11}$$
 

Monoxdifly

Well-known member
Aug 6, 2015
271
I think I get it now. Thanks to both of you.