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Sum to n terms of the series

 tan113+tan117+tan1113+is 

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a
tan−1⁡nn+2
b
tan−1⁡2n−12n+1
c
tan−1⁡13n
d
tan−1⁡12n

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detailed solution

Correct option is A

ar=tan−1⁡11+r(r+1)=tan−1⁡r+1−r1+(r+1)r=tan−1⁡(r+1)−tan−1⁡(r)Thus , Sn=∑r=1n ar=tan−1⁡(n+1)−tan−1⁡(1)=tan−1⁡n+1−11+(n+1)(1)=tan−1⁡nn+2


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The sum of the first n  terms of the series

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