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Questions  

limn1n1+n2n2+12+n2n2+22++n2n2+(n1)2

is equal to

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a
π2
b
π3
c
π4
d
π6

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detailed solution

Correct option is C

limn→∞ 1n1+n2n2+12+n2n2+22+⋯+n2n2+(n−1)2=limn→∞ 1n∑r=0n−1 n2n2+r2=limn→∞ 1n∑r=0n−1 11+rn2=∫01 dx1+x2=tan−1 x01=π4


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