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Q.

limn→∞ 1n1+n2n2+12+n2n2+22+⋯+n2n2+(n−1)2is equal to

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a

π2

b

π3

c

π4

d

π6

answer is C.

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Detailed Solution

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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