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Find the sum of first \(9\) terms of the series.
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\(\Large \color{black }{\begin{align} \dfrac{1^3}{1}+\dfrac{1^3+2^3}{1+3}+\dfrac{1^3+2^3+3^3}{1+3+5}+\cdots \hspace{.33em}\\~\\ \end{align}}\)
both numerator and denominator sums have formulas : \[\dfrac{\left(\dfrac{n(n+1)}{2}\right)^2}{n^2}\] simplifying gives you \(\dfrac{(n+1)^2}{4}\)
quick!
congrats you're green! i thought id get green first as you're stuck at 89 like forever haha
haha yes nervous nineties!
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