If frac{1^3+2^3+3^3+ldots text{ upto } n text | vedclass

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(D) The sum of the numerator is $\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2$.The denominator is $\sum_{r=1}^n r(2r+1) = \sum_{r=1}^n (2r^2+r)

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$5$

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(D) The sum of the numerator is $\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2$.The denominator is $\sum_{r=1}^n r(2r+1) = \sum_{r=1}^n (2r^2+r) = 2\sum r^2 + \sum r$.Using standard summation formulas: $2 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6} [2(2n+1) + 3] = \frac{n(n+1)(4n+5)}{6}$.Given the ratio: $\frac{\frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(4n+5)}{6}} = \frac{9}{5}$.Simplifying: $\frac{n(n+1)}{4} \cdot \frac{6}{4n+5} = \frac{3n(n+1)}{2(4n+5)} = \frac{9}{5}$.$\frac{n(n+1)}{2(4n+5)} = \frac{3}{5} \Rightarrow 5n^2 + 5n = 6(4n+5) = 24n + 30$.$5n^2 - 19n - 30 = 0$.$(n-5)(5n+6) = 0$.Since $n$ must be a positive integer,$n = 5$.

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

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