lim n → ∞ n n 2 + 1 2 + n n 2 + 2 2 + n n 2 + 3 2 + . .   … . + 1 5 n 2 is equal to

limnnn2+12+nn2+22+nn2+32+.. .+15n2 is equal to
  1. π4
  2. tan-12
  3. π2
  4. tan-1(3)

Solution

To find,

limnnn2+12+nn2+22++nn2+2n2

=r=12nnn2+r2=1nr=12n11+rn2  

=0211+x2dx=tan-1x02

=tan-12.

Asked in: JEE Main 2019 (12 Jan Shift 2)

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