Verified Solution • 2025–26 Session moderate QJEEM23Ma6085: JEE Main 2023 Mathematics - Class 12 JEEM23Ma6085 Solution The value of limn→∞1+2−3+4+5−6 + ..... + (3n−2)+(3n−1)−3n2n4+4n+3−n4+5n+4\mathop {\lim }\limits_{n \to \infty } {{1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n} \over {\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4} }}n→∞lim2n4+4n+3−n4+5n+41+2−3+4+5−6+.....+(3n−2)+(3n−1)−3n is : Multiple Choice Options A 322{3 \over {2\sqrt 2 }}223 B 32(2+1){3 \over 2}(\sqrt 2 + 1)23(2+1) C 3(2+1)3(\sqrt 2 + 1)3(2+1) D 2+12{{\sqrt 2 + 1} \over 2}22+1 Step-by-Step Verified Solution amp;I=limn→∞(1+2+3+…+3n)−2(3+6+9+..+3n)2n4+4n+3−n4+5n+4amp;=limn→∞3n(3n+1)2−6n(n+1)2(2n4+4n+3−n4+5n+4)amp;=limn→∞3n(n−1)[2n4+4n+3+n4+5n+4]2⋅[(2n4+4n−3)−(n4+5n+4)]amp;=limn→∞3⋅1⋅(1−1n)[2+4n3+3n4+1+5n3+4n4]2[1−1n3−7n4]amp;=3(2+1)2amp;\begin{aligned}& I=\lim _{n \rightarrow \infty} \frac{(1+2+3+\ldots+3 n)-2(3+6+9+. .+3 n)}{\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}} \\\\& =\lim _{n \rightarrow \infty} \frac{\frac{3 n(3 n+1)}{2}-6 \frac{n(n+1)}{2}}{\left(\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}\right)} \\\\& =\lim _{n \rightarrow \infty} \frac{3 n(n-1)\left[\sqrt{2 n^4+4 n+3}+\sqrt{n^4+5 n+4}\right]}{2 \cdot\left[\left(2 n^4+4 n-3\right)-\left(n^4+5 n+4\right)\right]} \\\\& =\lim _{n \rightarrow \infty} \frac{3 \cdot 1 \cdot\left(1-\frac{1}{n}\right)\left[\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right]}{2\left[1-\frac{1}{n^3}-\frac{7}{n^4}\right]} \\\\& =\frac{3(\sqrt{2}+1)}{2} \\&\end{aligned}amp;I=n→∞lim2n4+4n+3−n4+5n+4(1+2+3+…+3n)−2(3+6+9+..+3n)amp;=n→∞lim(2n4+4n+3−n4+5n+4)23n(3n+1)−62n(n+1)amp;=n→∞lim2⋅[(2n4+4n−3)−(n4+5n+4)]3n(n−1)[2n4+4n+3+n4+5n+4]amp;=n→∞lim2[1−n31−n47]3⋅1⋅(1−n1)[2+n34+n43+1+n35+n44]amp;=23(2+1)amp; Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring