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Series Problem | PRMO 2017 | Question 6

Try this beautiful problem from the Pre-RMO, 2017 based on Series.

Series Problem - PRMO 2017


Let the sum \displaystyle\sum_{n=1}^{9}\frac{1}{n(n+1)(n+2)} written in its lowest terms be \frac{p}{q}, find the value of q-p.

  • is 107
  • is 83
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Integers

Rearrangement of terms in series

Check the Answer


Answer: is 83.

PRMO, 2017, Question 6

Calculus Vol 1 and 2 by Apostle

Try with Hints


here \displaystyle\sum_{n=1}^{9}\frac{n+2-n}{2n(n+1)(n+2)}

=\frac{1}{2}\displaystyle\sum_{n=1}^{9}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})

=\frac{1}{2}(\frac{1}{(1)(2)}-\frac{1}{(2)(3)}

+\frac{1}{(2)(3)}-\frac{1}{(3)(4)}+....+\frac{1}{(9)(10)}-\frac{1}{(10)(11)})

=\frac{1}{2}(\frac{1}{2}-\frac{1}{110})

\Rightarrow \frac{27}{110}

\Rightarrow q-p=110-27

=83.

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Try this beautiful problem from the Pre-RMO, 2017 based on Series.

Series Problem - PRMO 2017


Let the sum \displaystyle\sum_{n=1}^{9}\frac{1}{n(n+1)(n+2)} written in its lowest terms be \frac{p}{q}, find the value of q-p.

  • is 107
  • is 83
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Integers

Rearrangement of terms in series

Check the Answer


Answer: is 83.

PRMO, 2017, Question 6

Calculus Vol 1 and 2 by Apostle

Try with Hints


here \displaystyle\sum_{n=1}^{9}\frac{n+2-n}{2n(n+1)(n+2)}

=\frac{1}{2}\displaystyle\sum_{n=1}^{9}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})

=\frac{1}{2}(\frac{1}{(1)(2)}-\frac{1}{(2)(3)}

+\frac{1}{(2)(3)}-\frac{1}{(3)(4)}+....+\frac{1}{(9)(10)}-\frac{1}{(10)(11)})

=\frac{1}{2}(\frac{1}{2}-\frac{1}{110})

\Rightarrow \frac{27}{110}

\Rightarrow q-p=110-27

=83.

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