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$$\begin{gathered} To solve : The value of k for which 1^3+2^3+3^3+4^3+...+k^3=2025 \\ Solution : 1^3+2^3+3^3+4^3+...+k^3=2025 \\ \Rightarrow \sum _{i=1}^k{i}^3=2025 \\ and it is given by the formula \\ \Rightarrow {\left\{\frac{k\left(k+1\right)}{2}\right\}}^2=2025 \\ taking square root both the sides \\ \Rightarrow \sqrt{{\left\{\frac{k\left(k+1\right)}{2}\right\}}^2}=\sqrt{2025} \\ \Rightarrow \frac{k\left(k+1\right)}{2}=45 \\ \Rightarrow k(k+1)=45\times 2 \\ \Rightarrow k^2+k=90 \\ \Rightarrow k^2+k-90=0 \\ \Rightarrow k^2+10k-9k-90=0 \\ \Rightarrow k(k+10)-9(k+10)=0 \\ \Rightarrow (k+10)(k-9)=0 \\ either k=-10 or k=9 \\ as we can see k cannot be negative \\ so k=9 \end{gathered}$$
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