If 1 3 + 2 3 + 3 3 + . . . u p t o n t e r m s 1 · 3 + 2 · 5 + 3 · 7 + . . . u p t o n t e r m s = 9 5 , then the value of n is _______. [2023]

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Solution

Q.
If $\frac{1^{3} + 2^{3} + 3^{3} + . . . u p t o n t e r m s}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + . . . u p t o n t e r m s} = \frac{9}{5},$ then the value of $n$ is _______. [2023]

Ans.
(5)

$Given,$ $\frac{1^{3} + 2^{3} + 3^{3} + \dots up to n terms}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots up to n terms} = \frac{9}{5}$

$\Rightarrow \frac{{(\frac{n (n + 1)}{2})}^{2}}{\sum_{r = 1}^{n} r (2 r + 1)} = \frac{9}{5}$ $\Rightarrow \frac{{(\frac{n (n + 1)}{2})}^{2}}{2 \sum_{r = 1}^{n} r^{2} + \sum_{r = 1}^{n} r} = \frac{9}{5}$

$\Rightarrow \frac{{(\frac{n (n + 1)}{2})}^{2}}{2 (\frac{n (n + 1) (2 n + 1)}{6}) + \frac{n (n + 1)}{2}} = \frac{9}{5}$

$\Rightarrow \frac{{[\frac{n (n + 1)}{2}]}^{2}}{\frac{n (n + 1)}{2} [\frac{2 (2 n + 1)}{3} + 1]} = \frac{9}{5} \Rightarrow \frac{\frac{n (n + 1)}{2}}{\frac{4 n + 5}{3}} = \frac{9}{5}$

$\Rightarrow 5 n^{2} - 19 n - 30 = 0 \Rightarrow (5 n + 6) (n - 5) = 0 \Rightarrow n = \frac{- 6}{5}, 5$

$Hence, n = 5$

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