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JEE Main 2017 · Mathematics · Sequences and Series · medium

The value of (1)/(1·4)+(1)/(4·7)+(1)/(7·10)+…+(1)/(25·28) is

Single correct option

The value of 11⋅4+14⋅7+17⋅10+⋯+125⋅28\dfrac1{1\cdot4}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dots+\dfrac1{25\cdot28} is
  1. A. 2728\dfrac{27}{28}
  2. B. 928\dfrac{9}{28}
  3. C. 328\dfrac{3}{28}
  4. D. 14\dfrac{1}{4}
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Answer: (B) 928\dfrac{9}{28}

The general term is 1(3k+1)(3k+4)\dfrac1{(3k+1)(3k+4)} for k=0,1,…,8k=0,1,\dots,8 (nine terms). Using partial fractions, 1(3k+1)(3k+4)=13(13k+1−13k+4)\dfrac1{(3k+1)(3k+4)}=\dfrac13\left(\dfrac1{3k+1}-\dfrac1{3k+4}\right). The sum telescopes: 13(11−128)=13⋅2728=928\dfrac13\left(\dfrac11-\dfrac1{28}\right)=\dfrac13\cdot\dfrac{27}{28}=\dfrac{9}{28}. Forgetting the factor 13\tfrac13 gives the trap value 2728\tfrac{27}{28}.
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