JEE Main 2021MathematicsLimitsHardMCQ

JEE Main 2021 — Limits Question with Solution

JEE Main 2021 (26 Aug Shift 2)

Question

limx→2∑n=19xn(n+1)x2+2(2n+1)x+4 is equal to :

Choose an option

Show full solutionCorrect option: D
Correct answer
D944

Step-by-step explanation

limx→2  ∑n=19xn(n+1)x2+2(2n+1)x+4

limx→2∑  n=19xnn+1x2+2nx+2n+1x+4

limx→2∑  n=19(n+1)x+2-nx+2(n+1)x+2nx+2

limx→2∑  n=191(nx+2)-1(n+1)x+2

=1x+2-12x+2

12x+2-13x+2

....+19x+2-110x+21x+2-110x+2

⇒  limx→21x+2-110x+2=944

⇒  limx→210x+2-x-2x+210x+2

limx→29xx+210x+2=184×22=944

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About this question

This is a previous-year question from JEE Main 2021, covering the Limits chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.