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Consider a sequence of numbers $a_1, a_2, a_3, \dots , a_n$ where $a_n = \frac{1}{n}-\frac{1}{n+2}$, for each integer $n>0$. Whart is the sum of the first 50 terms?

  1. $\left( 1+ \frac{1}{2} \right) - \frac{1}{50}$
  2. $\left( 1+ \frac{1}{2} \right) + \frac{1}{50}$
  3. $\left( 1+ \frac{1}{2} \right) - \left( \frac{1}{51} + \frac{1}{52} \right)$
  4. $1 - \left( \frac{1}{51} + \frac{1}{52} \right)$
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Migrated from GO Civil 1 year ago by Arjun
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Answer is C

1 Answer

2 votes
 
Best answer

Sequence of numbers $a_{1},a_{2},a_{3},\ldots,a_{n}$

where $a_{n}=\frac{1}{n}-\frac{1}{n+2}$  for each  integer $n>0$

Now we find the $1^{st}$ term, $2^{nd}$ term up to $50^{th}$ term

$a_{1}=\frac{1}{1}-\frac{1}{1+2}=1-\frac{1}{3}$

$a_{2}=\frac{1}{2}-\frac{1}{2+2}=\frac{1}{2}-\frac{1}{4}$

$a_{3}=\frac{1}{3}-\frac{1}{3+2}=\frac{1}{3}-\frac{1}{5}$

$a_{3}=\frac{1}{4}-\frac{1}{4+2}=\frac{1}{4}-\frac{1}{6}$
 


$a_{50}=\frac{1}{50}-\frac{1}{50+2}=\frac{1}{50}-\frac{1}{52}$

Sum of the first $50$ terms

$S=a_{1}+a_{2}+a_{3}+a_{4}+\ldots+a_{50}$

$S=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\ldots+\frac{1}{50}-\frac{1}{52}$

We can write like this
$S=\left (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…..+\frac{1}{50} \right )-\left ( \frac{1}{3}+\frac{1}{4}+\frac{1}{5} +\frac{1}{6}+\ldots+\frac{1}{52}\right )$
$\quad=\left (1+\frac{1}{2}\right )+ \left \{\left (\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{50} \right ) -\left ( \frac{1}{3}+\frac{1}{4}+\frac{1}{5} +\frac{1}{6}+\ldots+\frac{1}{50}\right ) \right \}-\frac{1}{51}-\frac{1}{52}$
$\quad=\left ( 1+\frac{1}{2}\right )-\left (\frac{1}{51}+\frac{1}{52} \right )$

So, correct options is (C).


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