JEE Main 16 March 2021 Shift 1 Solved Paper

Section: Mathematics

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Question : 80 of 90

Marks: +1, -0

Let Sk=

k

∑

r=1

tan−1(‌

6r

22r+1+32r+1

). Then,

lim

k→∞

Sk is equal to

[16 Mar 2021 Shift 1]

tan−1(‌
3
2
)
‌
π
2
cot−1(‌
3
2
)
tan−1(3)

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