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Q.

Let P(k)=1+cos⁡π4k1+cos⁡(2k−1)π4k 1+cos⁡(2k+1)π4k1+cos⁡(4k−1)π4k. Then

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a

P(3)=116

b

P(4)=2−216

c

P(5)=3−532

d

P(6)=2−316

answer is A.

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Detailed Solution

P(k)=1+cos⁡π4k1+cos⁡π2−π4k1+cos⁡π2+π4k1+cos⁡π−π4k=1+cos⁡π4k1+sin⁡π4k1−sin⁡π4k1−cos⁡π4k=1−cos2⁡π4k1−sin2⁡π4k=4sin2⁡π4k⋅cos2⁡π4k4P(k)=14sin2⁡π2k⇒P(3)=14×14=116⇒P(4)=π4sin2⁡π2k=14sin2⁡π8=181−cos⁡π4=2−216⇒P(5)=14sin2⁡π10=182sin2⁡π10=181−cos⁡36∘=181−5+14=3−532⇒P(6)=14sin2⁡π12=182sin2⁡π12=181−cos⁡π6⇒P6=2−316
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