\(8\sqrt2^n\)
\(2^{(n+2)}\sqrt2\)
\(\sqrt2^{(n+3)}\)
\(8n\sqrt2\)
Correct answer is B
8√2, 16√2, 32√2, ..
\(a = 8\sqrt2; r =\frac{T_2}{T_1}=\frac{16\sqrt2}{8\sqrt2}=2\)
\(T_n=ar^{n-1}\)
\(T_n=8\sqrt2 \times 2^{n-1}\)
\(T_n=2^3\times2^{n-1}\times\sqrt2\)
\(T_n=2^{3+n-1}\times\sqrt2\)
\(\therefore T_n= 2^{(n+2)}\sqrt2\)