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.

Other useful links

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.

Other useful links

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