Perfect Square from Geometric Progression

Question

This question is from QuantGuide(namely Geometrical Progression):
Write out a series of whole numbers in geometrical progression with at least 3 terms, starting from 1, so that the numbers add up to a square. The common ratio must be strictly larger than 1.
Give the answer in the form of the smallest square number in which a progression can be written.
My Approach:
I have arrived at this position \begin{equation} 1+r+r^2+r^3+...+r^{n-1} = b^2 \end{equation} where r,n,b are some integers and $r>1$. But I can't proceed further. Hints would be appreciated.

Answer 1

Smallest square number is $121$

$$1 + 3 + 9 + 27 + 81 = 121$$

Proof: Try perfect square numbers less than $121$ and see that you cannot write them that way.

This question relates to consecutive powerful numbers, a concept not yet fully understood. (because $r^n$ is powerful and $r^n-1$ is potentially a powerful number)

There is also an open question: "Can three consecutive numbers be powerful?"

Answer 2

There are only two square numbers which can be expressed as a sum of consecutive powers of integers. These are $121$ and $400$.

$$1+3+9+27+81=121$$

$$1+7+49+343 = 400$$