A natural number k>1 is called good if there exist natural numbers...

Let $\mathrm{a}_1=\mathrm{A}_1^2, \mathrm{a}_2=\mathrm{A}_2^2 \cdots \cdots \cdots \cdots \cdot \mathrm{a}_{\mathrm{k}}=\mathrm{A}_{\mathrm{k}}^2$ we have to check if it is possible for distinct natural number $\mathrm{A}_1, \mathrm{~A}_2$. $\_\_\_\_$ $\mathrm{A}_{\mathrm{k}}$ to satisfy, $\frac{1}{A_1}+\frac{1}{A_2}+\ldots \ldots . \frac{1}{A_k}=1$ For $k=2$; it is obvious that there do not exist distinct $A_1, A_2$, such that $\frac{1}{A_1}+\frac{1}{A_2}=1 \Rightarrow 2$ is not a good number. For $\mathrm{k}=3$; we have $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1 \Rightarrow 3$ is a good number. $\frac{1}{2}+\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=1 \Rightarrow 4$ is a good number. $k$ will be a good numbers for all $k \geq 3$ $$\begin{aligned} & f(n)=3+4+\ldots n \text { terms }=\frac{n(n+5)}{2} ; f(n+5)=\frac{(n+5)(n+10)}{2} \\ & \frac{f(n+5)}{f(n)}=\frac{n+10}{n}=1+\frac{10}{n} \end{aligned}$$ Which will be integer for $\mathrm{n}=1,2,5$ and 10 $\text { sum }=1+2+5+10=18$