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GATE 2018 CE (CE1) – Question 9

General Aptitude · Quantitative Aptitude: Algebra, Geometry and Mensuration · 2 marks · Multiple choice

For each integer $n>0$, $a_n=1/n-1/(n+2)$. Find the sum of the first 50 terms.

  1. $(1+1/2)-1/50$
  2. $(1+1/2)+1/50$
  3. $(1+1/2)-(1/51+1/52)$
  4. $1-(1/51+1/52)$

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Show answer and explanation

Correct answer: (C) $(1+1/2)-(1/51+1/52)$

Explanation

Expand the sum and cancel matching terms: $$\sum_{n=1}^{50}\left(\frac1n-\frac1{n+2}\right)=1+\frac12-\frac1{51}-\frac1{52}.$$ Answer C.