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Q.

Let Sk=∑r=1ktan-16r22r+1+32r+1, the value of limk→∞Sk is equal to¯¯ :

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a

tan−132

b

π2

c

cot−132

d

tan−13

answer is C.

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Detailed Solution

SK=∑r=1Ktan−13r2r+11+322r+1=∑  tan−132r+1−32r1+32r+1.32r=∑tan−132r+1−tan−132r=π2−tan−132=cot−132
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