Question

If $\frac{1^3+2^3+3^3+\ldots \text { up to } n \text { terms }}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \text { up to } n \text { terms }}=\frac{9}{5}$, then the value of $n$ is

Answer 1

The correct answer is 5
13+23+33….⋅+n3=(2n(n+1)​)2
1⋅3+2⋅5+3⋅7+……+n terms =
r=1∑n​r(2r+1)=r=1∑n​(2r2+r)
=62⋅n(n+1)(2n+1)​+2n(n+1)​
=6n(n+1)​(2(2n+1)+3)
=2n(n+1)​×3(4n+5)​
=2n(n+1)​×3(4n+5)​4n2(n+1)2​​=59​
⇒25n(n+1)​=39(4n+5)​
⇒15n(n+1)=18(4n+5)
⇒15n2+15n=72n+90
⇒15n2−57n−90=0⇒5n2−19n−30=0
⇒(n−5)(5n+6)=0
⇒n=5−6​ or 5
⇒n=5.