LimitsmediumFree

Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
lim⁡n→∞12n+22(n−1)+32(n−2)+…+n2⋅113+23+33+…+n3\displaystyle\lim_{n\to\infty}\dfrac{1^{2}n+2^{2}(n-1)+3^{2}(n-2)+\ldots+n^{2}\cdot 1}{1^{3}+2^{3}+3^{3}+\ldots+n^{3}} equal
A13\dfrac{1}{3}correct
B23\dfrac{2}{3}
C12\dfrac{1}{2}
D16\dfrac{1}{6}
Solution
Step 1: Numerator general term: r2(n−r+1)=r2(n+1)−r3r^{2}(n-r+1)=r^{2}(n+1)-r^{3}.
N=∑r=1nr2(n−r+1)=(n+1)∑r2−∑r3N=\sum_{r=1}^{n}r^{2}(n-r+1)=(n+1)\sum r^{2}-\sum r^{3}
 Step 2: Use ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n}r^{2}=\dfrac{n(n+1)(2n+1)}{6} and ∑r=1nr3=(n(n+1)2)2\sum_{r=1}^{n}r^{3}=\left(\dfrac{n(n+1)}{2}\right)^{2}:
N=(n+1)⋅n(n+1)(2n+1)6−n2(n+1)24N=(n+1)\cdot\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n^{2}(n+1)^{2}}{4}
=n(n+1)2(2n+1)6−n2(n+1)24=n(n+1)212[2(2n+1)−3n]=\dfrac{n(n+1)^{2}(2n+1)}{6}-\dfrac{n^{2}(n+1)^{2}}{4}=\dfrac{n(n+1)^{2}}{12}[2(2n+1)-3n]
=n(n+1)2(n+2)12=\dfrac{n(n+1)^{2}(n+2)}{12}
 Step 3: Denominator: D=n2(n+1)24D=\dfrac{n^{2}(n+1)^{2}}{4}. Step 4: Ratio:
ND=n(n+1)2(n+2)/12n2(n+1)2/4=4(n+2)12n=n+23n\dfrac{N}{D}=\dfrac{n(n+1)^{2}(n+2)/12}{n^{2}(n+1)^{2}/4}=\dfrac{4(n+2)}{12n}=\dfrac{n+2}{3n}
 Step 5: Take n→∞n\to\infty:
lim⁡n→∞n+23n=13\lim_{n\to\infty}\dfrac{n+2}{3n}=\dfrac{1}{3}
 Correct answer: (1) 13\dfrac{1}{3}
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Limits · easy
x x ( x+1 - x ) equal to
Limits · easy
x r=1 10 (x+r) 2010 (x 1006 +1)(2x 1004 +1) , is equal to
Limits · easy
x 0 ae x +b x+ce -x e 2x -2e x +1 =4 . Then
Limits · easy
The value of x 1 x 1/3 -x 1/4 x 3 -1 is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath