Sequences & Series
Sum of cubes vs sum of products
nta_pyq_2023_jan
Grade 11
Question:
If $\dfrac{1^3 + 2^3 + 3^3 + \ldots \text{ upto n terms}}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \ldots \text{ upto n terms}} = \dfrac{9}{5}$, then the value of n is ______.
Step-by-Step Solution
Key Concept: Numerator = $\left(\frac{n(n+1)}{2}\right)^2$. Denominator = $\sum_{r=1}^n r(2r+1) = \frac{n(n+1)(4n+5)}{6}$. Set their ratio equal to $\frac{9}{5}$.
$\dfrac{\frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(4n+5)}{6}} = \dfrac{9}{5} \Rightarrow \dfrac{3n(n+1)}{2(4n+5)} = \dfrac{9}{5} \Rightarrow 15n(n+1) = 18(4n+5) \Rightarrow 5n^2 - 19n - 30 = 0 \Rightarrow (n-5)(5n+6)=0 \Rightarrow n=5$.
Correct Answer: 5