Given cot−1(α)=cot−1(2)+cot−1(8)+cot−1(18)+….
=n=1∑100tan−1(4n22)
=n=1∑100tan−1(1+(2n+1)(2n−1)(2n+1)−(2n−1))
=n=1∑100tan−1(2n+1)−tan−1(2n−1)
=tan−1201−tan−11
=tan−1(202200)
∴cot−1(α)=cot−1(200202)
α=1.01
If cot−1(α)=cot−12+cot−18+cot−118+cot−132+…. upto 100 terms, then α is:
Held on 17 Mar 2021 · Verified 6 Jul 2026.
1.01
1.00
1.02
1.03
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