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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Definite Integration

NDA 2016 · Mathematics · Definite Integration

Given that a_ n =∫_0^π ^2\ (n+1) x\ 2 x dx What is a_ n-1 -a_ n-4...

Given that an=∫0πsin⁡2{(n+1)x}sin⁡2xdx\mathrm{a}_{\mathrm{n}}=\int_0^\pi \frac{\sin ^2\{(\mathrm{n}+1) \mathrm{x}\}}{\sin 2 \mathrm{x}} \mathrm{dx} What is an−1−an−4a_{n-1}-a_{n-4} equal to ?

Options

  1. A–1
  2. B0
  3. C1
  4. D2

Correct answer

B. 0

Step-by-step solution

∵\because given sequence ana_n also AP with no difference. Thus an−1−an−4=0a_{n-1}-a_{n-4}=0
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