Ответ: b₉=2N+1, b₁₀=(√(2N+1))³

Объяснение:

[tex]\displaystyle\\b_1=\frac{1}{(2N+1)^3}\ \ \ \ \ \ q=\sqrt{2N+1} \ \ \ \ \ \ b_8=?\\\\b_9=b_1q^{9-1}= b_1q^{8}= \frac{1}{(2N+1)^3} *(\sqrt{2N+1)})^8=\frac{(2N+1)^4}{(2N+1)^3} =2N+1.[/tex]

[tex]b_{10}=b_9*q=(2N+1)*\sqrt{2N+1} =(\sqrt{ 2N+1})^3.[/tex]