JEE Main 2023 — Mathematics Algebra

Given,

$$\frac{{1}^{3}+{2}^{3}+{3}^{3}.......{n}^{3}}{1\cdot 3+2\cdot 5+3\cdot 7+....n(2n+1)}=\frac{9}{5}$$1⋅3+2⋅5+3⋅7+....n(2n+1)13+23+33.......n3​=59​

Now we know that ${1}^{3}+{2}^{3}+{3}^{3}.......{n}^{3}={[\frac{n(n+1)}{2}]}^{2}$13+23+33.......n3=[2n(n+1)​]2

And $\Sigma n(2n+1)=2\Sigma {n}^{2}+\Sigma n$Σn(2n+1)=2Σn2+Σn

$$\Rightarrow \Sigma n(2n+1)=2\times \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$$⇒Σn(2n+1)=2×6n(n+1)(2n+1)​+2n(n+1)​

$$\Rightarrow \Sigma n(2n+1)=\frac{n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}$$⇒Σn(2n+1)=3n(n+1)(2n+1)​+2n(n+1)​

So putting the value in the given equation we get,

$$\frac{{1}^{3}+{2}^{3}+{3}^{3}.......{n}^{3}}{1\cdot 3+2\cdot 5+3\cdot 7+....n(2n+1)}=\frac{9}{5}$$1⋅3+2⋅5+3⋅7+....n(2n+1)13+23+33.......n3​=59​

$$\Rightarrow \frac{{[\frac{n(n+1)}{2}]}^{2}}{\frac{n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}}=\frac{9}{5}$$⇒3n(n+1)(2n+1)​+2n(n+1)​[2n(n+1)​]2​=59​

$$\Rightarrow \frac{\frac{1}{4}[n(n+1)]}{\frac{(2n+1)}{3}+\frac{1}{2}}=\frac{9}{5}$$⇒3(2n+1)​+21​41​[n(n+1)]​=59​

$$\Rightarrow \frac{6[n(n+1)]}{4(2(2n+1)+3)}=\frac{9}{5}$$⇒4(2(2n+1)+3)6[n(n+1)]​=59​

$$\Rightarrow \frac{[n(n+1)]}{2(2(2n+1)+3)}=\frac{3}{5}$$⇒2(2(2n+1)+3)[n(n+1)]​=53​

$$\Rightarrow 5[n(n+1)]=6(2(2n+1)+3)$$⇒5[n(n+1)]=6(2(2n+1)+3)

$$\Rightarrow 5{n}^{2}+5n=6(4n+5)$$⇒5n2+5n=6(4n+5)

$$\Rightarrow 5{n}^{2}+5n=24n+30$$⇒5n2+5n=24n+30

$$\Rightarrow 5{n}^{2}-19n-30=0$$⇒5n2−19n−30=0

$$\Rightarrow 5{n}^{2}-25n+6n-30=0$$⇒5n2−25n+6n−30=0

$$\Rightarrow (5n+6)(n-5)=0$$⇒(5n+6)(n−5)=0

$$\Rightarrow n=5$$⇒n=5