JEE Main 2023 — Mathematics Calculus

The given limit $\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{1+2-3+4+5-6+\ldots +(3n-2)+(3n-1)-3n}{\sqrt{2{n}^{4}+4n+3-}\sqrt{{n}^{4}+5n+4}}$n→∞lim​2n4+4n+3−​n4+5n+4​1+2−3+4+5−6+…+(3n−2)+(3n−1)−3n​ can be expressed as,$\displaystyle \underset{n\rightarrow \infty }{\mathrm{lim}}\frac{\sum _{r=1}^{n}(3r-2)+(3r-1)-3r}{\sqrt{2{n}^{4}+4n+3}-\sqrt{{n}^{4}+5n+4}}$n→∞lim​2n4+4n+3​−n4+5n+4​∑r=1n​(3r−2)+(3r−1)−3r​

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{\sum _{r=1}^{n}3(r-1)}{\sqrt{2{n}^{4}+4n+3}-\sqrt{{n}^{4}+5n+4}}$$=n→∞lim​2n4+4n+3​−n4+5n+4​∑r=1n​3(r−1)​

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{0+3+6+9+\ldots .nterms}{\sqrt{2{n}^{4}+4n+3}-\sqrt{{n}^{4}+5n+4}}.......(1)$$=n→∞lim​2n4+4n+3​−n4+5n+4​0+3+6+9+….nterms​.......(1)

We know the sum of the nth term of A.P. is given by

$${S}_{n}=\frac{n}{2}[2a+(n-1)d]$$Sn​=2n​[2a+(n−1)d]

Using above formula in equation(1) we get,

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{\frac{3n(n-1)}{2}}{(\sqrt{2{n}^{4}+4n+3}-\sqrt{{n}^{4}+5n+4})}$$=n→∞lim​(2n4+4n+3​−n4+5n+4​)23n(n−1)​​

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{3n(n-1)}{2(\sqrt{2{n}^{4}+4n+3}-\sqrt{{n}^{4}+5n+4})}$$=n→∞lim​2(2n4+4n+3​−n4+5n+4​)3n(n−1)​

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{3(1-\frac{1}{n})}{2(\sqrt{2+\frac{4}{{n}^{3}}+\frac{3}{{n}^{4}}}-\sqrt{1+\frac{5}{{n}^{3}}+\frac{4}{{n}^{4}}})}$$=n→∞lim​2(2+n34​+n43​​−1+n35​+n44​​)3(1−n1​)​

$$=\frac{3(1-0)}{2(\sqrt{2+0+0}-\sqrt{1+0+0})}$$=2(2+0+0​−1+0+0​)3(1−0)​

$$=\frac{3}{2(\sqrt{2}-1)}=\frac{3}{2}(\sqrt{2}+1)$$=2(2​−1)3​=23​(2​+1)