Site navigation

The sum up to \(10\) terms of the series \(1 \cdot 3 + 5 \cdot 7 + 9 \cdot 11 + ...\) is

Question and options

The sum up to 1010 terms of the series 1⋅3+5⋅7+9⋅11+...1 \cdot 3 + 5 \cdot 7 + 9 \cdot 11 + ... is
  1. 4590
  2. 5310
  3. 4930
  4. 3867

Answer and explanation

Correct answer: B - 5310

The sum is 53105310. With k=0,…,9k=0,\ldots,9, each product is (4k+1)(4k+3)=16k2+16k+3(4k+1)(4k+3)=16k^2+16k+3, so the sum is 16(285)+16(45)+30=531016(285)+16(45)+30=5310. Dropping the linear term gives 45904590.

Practice for these exams

  • IPMAT Indore

Related question papers

Related exam syllabi

More solved explanations in Number series & simplification