lim n → ∞ 1 1 + n 5 + 2 4 2 5 + n 5 + 3 4 3 5 + n 5 + … + n 4 n 5 + n 5 =

limn11+n5+2425+n5+3435+n5++n4n5+n5=
  1. 15log3
  2. 13log5
  3. 12log5
  4. log25

Solution

limn11+n5+2425+n5+3435+n5++n4n5+n5

=limnr=1nr4r5+n5

=limn1nr=1nrn4rn5+1=01x41+x5dx

=15ln1+x501=15ln2=ln25

Asked in: AP EAMCET 2022 (04 Jul Shift 1)

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