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📐 Mathematics  ·  Sequences and Series  ·  JEE

Sum: 1/(1×2×3) + 1/(2×3×4) + ... to n terms =

Answer: 1/4 - 1/(2(n+1)(n+2)).

  • A 1/4 - 1/(2(n+1)(n+2))
  • B 1/4, without the correction term for finite n
  • C 1/2(n+1), a partial telescoping result
  • D n/(n+1)(n+2), a related but different ratio

Correct answer: A. 1/4 - 1/(2(n+1)(n+2))

Explanation: Using partial fractions: each term = 1/2 [1/(k(k+1)) - 1/((k+1)(k+2))]. Telescoping: 1/2 [1/(1×2) - 1/((n+1)(n+2))] = 1/4 - 1/(2(n+1)(n+2)).

Each bar grows by the common difference d=3a1=2a2=5a3=8a4=11d=3

Bar heights for AP 2, 5, 8, 11 with the common difference d=3 marked between consecutive terms.

Concept context

Arithmetic progressions, geometric progressions, and sums

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