• MHB
• Monoxdifly
In summary, the given expression can be rewritten as a series with a common denominator of 231. By summing the first few terms and looking for a pattern, the value of the series can be determined to be 10/11. Another method is to use algebraic manipulation to find the value directly, which also results in 10/11.
Monoxdifly
MHB
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 learned about sequence and series yet. Granted, the book also says that the question was from a middle school-level math contest.

Monoxdifly said:
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 learned 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 $$\displaystyle \sum_{n=2}^N \frac1{1+2+\ldots+n} = \frac{N-1}{N+1}.$$

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

I think I get it now. Thanks to both of you.

## What is the formula for adding fractions?

The formula for adding fractions is: (a/b) + (c/d) = (ad + bc) / (bd)

## How do you add fractions with different denominators?

To add fractions with different denominators, you first need to find the lowest common denominator (LCD) by finding the least common multiple (LCM) of the denominators. Then, you convert each fraction to an equivalent fraction with the LCD as the denominator. Finally, you can add the fractions using the formula mentioned above.

## Can you add fractions with different numerators?

Yes, you can add fractions with different numerators as long as the denominators are the same. If the denominators are different, you need to follow the steps mentioned above to find the LCD and convert the fractions before adding them.

## What is the result of adding 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + .... + 1/231?

The result of adding 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + .... + 1/231 is 77/210.

## Can you simplify the result of adding 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + .... + 1/231?

Yes, the result of 77/210 can be simplified to 11/30 by dividing both the numerator and denominator by their greatest common factor, which is 7 in this case.

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