[ASK] Find 1/(1×2)+1/(2×3)+1/(3×4)+…+1/(2009×2010)

  • MHB
  • Thread starter Monoxdifly
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In summary, the formula for finding the sum of a series is S = a/(1-r), where S is the sum, a is the first term, and r is the common ratio. In the specific series 1/(1×2)+1/(2×3)+1/(3×4)+…+1/(2009×2010), the first term is 1/2 and the common ratio is 1/(n×(n+1)), where n is the term number. There are 2009 terms in this series, with the last term having a value of approximately 4.975e-7. To simplify this series, it can be rewritten as (1/n) × (1/(n+
  • #1
Monoxdifly
MHB
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Does anyone know how to add these fractions?
\(\displaystyle \frac1{1\times2}+\frac1{2\times3}+\frac1{3\times4}+…+\frac1{2009\times2010}\)
I believe making them in \(\displaystyle \frac12+\frac1{6}+\frac1{12}+….+\frac1{421890}\) form isn’t the correct approach.
Is there anything we can cancel out?
 
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  • #2
Re: [ASK] Fracttion Addition

Monoxdifly said:
Does anyone know how to add these fractions?
\(\displaystyle \frac1{1\times2}+\frac1{2\times3}+\frac1{3\times4}+…+\frac1{2009\times2010}\)
I believe making them in \(\displaystyle \frac12+\frac1{6}+\frac1{12}+….+\frac1{421890}\) form isn’t the correct approach.
Is there anything we can cancel out?

What does Partial Fraction decomposition do for us? $\dfrac{1}{n\cdot(n+1)} = \dfrac{1}{n}-\dfrac{1}{n+1}$
 

FAQ: [ASK] Find 1/(1×2)+1/(2×3)+1/(3×4)+…+1/(2009×2010)

What is the formula for finding the sum of the series 1/(1×2)+1/(2×3)+1/(3×4)+…+1/(2009×2010)?

The formula for finding the sum of a series is S = a/(1-r), where S is the sum, a is the first term, and r is the common ratio. In this case, the first term is 1/(1×2) = 1/2 and the common ratio is 1/(n×(n+1)), where n is the term number. Therefore, the formula for this specific series is S = (1/2)/(1-1/(n×(n+1))).

How many terms are in the series 1/(1×2)+1/(2×3)+1/(3×4)+…+1/(2009×2010)?

There are 2009 terms in this series, as each term has a denominator that increases by 1 until it reaches 2010.

What is the value of the last term in the series 1/(1×2)+1/(2×3)+1/(3×4)+…+1/(2009×2010)?

The last term in the series is 1/(2009×2010), which has a value of approximately 4.975e-7.

How can I simplify this series to make it easier to calculate?

One way to simplify this series is to rewrite it as (1/n) × (1/(n+1)), which can then be simplified to 1/(n² + n). This can help with calculating the common ratio and sum formula.

What is the sum of the series 1/(1×2)+1/(2×3)+1/(3×4)+…+1/(2009×2010)?

The sum of the series is approximately 4.986, or 4.986e0, as calculated by plugging in 2009 for n in the simplified formula S = (1/2)/(1-1/(n×(n+1))).

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