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📐 Mathematics  ·  Sequences and Series  ·  JEE

Sum: 1/1×3 + 1/3×5 + 1/5×7 + ... to n terms =

Answer: n/(2n+1).

  • A n/(2n+1)
  • B n/(2n-1)
  • C 1/(2n+1)
  • D n/(n+1)

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

Explanation: Partial fractions (telescoping): 1/((2k-1)(2k+1)) = ½[1/(2k-1) − 1/(2k+1)]. Sum = ½[1 − 1/(2n+1)] = ½ · 2n/(2n+1) = n/(2n+1).

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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