Tip:A–D to answerE for explanationV for videoS to reveal answer
Let fn=∫02π(∑k=1nsink−1x)(∑k=1n(2k−1)sink−1x)cosxdx,n∈N. Then, f21−f20 is equal to ______.
- A.
41
(Correct Answer) - B.
40
- C.
42
- D.
43
Explanation
Solution:
-
Let sinx=t⟹cosxdx=dt. Limits 0 se 1.
-
fn=∫01(∑k=1ntk−1)(∑k=1n(2k−1)tk−1)dt.
-
Is expression ko solve karne par fn=n2 nikalta hai.
-
Final Result: f21−f20=212−202=441−400=41.
Explanation
Solution:
-
Let sinx=t⟹cosxdx=dt. Limits 0 se 1.
-
fn=∫01(∑k=1ntk−1)(∑k=1n(2k−1)tk−1)dt.
-
Is expression ko solve karne par fn=n2 nikalta hai.
-
Final Result: f21−f20=212−202=441−400=41.
