<s>[INST] Continued proportion implies $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$ I am trying to find a tricky way to proof these:

If $a,b,c,d$ are in continued proportion, prove that $$(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2 ,$$ This result could be extended to $$(a^2+b^2+c^2+d^2)(b^2+c^2+d^2+e^2)=(ab+bc+cd+de)^2$$ when  $a,b,c,d,e$ are in continued proportion.

The standard way for solving them could be putting $\frac{a}{b}=\frac{c}{d}=k$ then followed by substitution and which is followed by tedious algebraic manipulations,but that is not what I am looking for could these be solved in a less easy way using some other algebraic method/tricks? Please explain.
 [/INST] You know that $b=ka$, $c=kb$ etc so the lhs can be rewritten
$$(a^2+b^2+c^2)(k^2a^2+k^2b^2+k^2c^2) = k^2(a^2+b^2+c^2)^2$$
and the rhs can be written
$$(ka^2 + kb^2 + kc^2)^2 = k^2(a^2+b^2+c^2)^2$$
and you're done. The same trick works for the second example.
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