Try this beautiful problem from the Pre-RMO, 2017 based on Series.
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.
Series
Integers
Rearrangement of terms in series
But try the problem first...
Answer: is 83.
PRMO, 2017, Question 6
Calculus Vol 1 and 2 by Apostle
First hint
here \(\displaystyle\sum_{n=1}^{9}\frac{n+2-n}{2n(n+1)(n+2)}\)
=\(\frac{1}{2}\displaystyle\sum_{n=1}^{9}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})\)
Second Hint
=\(\frac{1}{2}(\frac{1}{(1)(2)}-\frac{1}{(2)(3)}\)
+\(\frac{1}{(2)(3)}-\frac{1}{(3)(4)}+....+\frac{1}{(9)(10)}-\frac{1}{(10)(11)})\)
=\(\frac{1}{2}(\frac{1}{2}-\frac{1}{110})\)
Final Step
\(\Rightarrow \frac{27}{110}\)
\(\Rightarrow q-p=110-27\)
=83.
Try this beautiful problem from the Pre-RMO, 2017 based on Series.
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.
Series
Integers
Rearrangement of terms in series
But try the problem first...
Answer: is 83.
PRMO, 2017, Question 6
Calculus Vol 1 and 2 by Apostle
First hint
here \(\displaystyle\sum_{n=1}^{9}\frac{n+2-n}{2n(n+1)(n+2)}\)
=\(\frac{1}{2}\displaystyle\sum_{n=1}^{9}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})\)
Second Hint
=\(\frac{1}{2}(\frac{1}{(1)(2)}-\frac{1}{(2)(3)}\)
+\(\frac{1}{(2)(3)}-\frac{1}{(3)(4)}+....+\frac{1}{(9)(10)}-\frac{1}{(10)(11)})\)
=\(\frac{1}{2}(\frac{1}{2}-\frac{1}{110})\)
Final Step
\(\Rightarrow \frac{27}{110}\)
\(\Rightarrow q-p=110-27\)
=83.
