Answer:
option A
Step-by-step explanation:
[tex]\huge\mathcal{Given}\\=>\sum_{n=3}^{12}20.\left(\frac{1}{2}\right)^{n-1}\\=>20\sum_{n=3}^{12}\left(\frac{1}{2}\right)^{n}\frac{1}{2^{-1}}\\=>40\sum_{n=3}^{12}\left(\frac{1}{2}\right)^{n}==>EQ00\\Let\:\:\:S=\sum_{n=3}^{12}\left(\frac{1}{2}\right)^{n}\\S=\frac{1}{2^3}+\frac{1}{2^4}+.......+\frac{1}{2^{11}}+\frac{1}{2^{12}}=>>EQ01\\S\:-\frac{1}{2^{12}}=\frac{1}{2^3}+\frac{1}{2^4}+.......+\frac{1}{2^{11}}==> EQ02\\
From\:\:\:EQ01\:\:\:\\S=\frac{1}{2^3}+\frac{1}{2}\left(\frac{1}{2^3}+\frac{1}{2^4}+.......+\frac{1}{2^{11}}\right)\\From\:\:EQ02\:\:\\S=\frac{1}{2^3}+\frac{1}{2}\left(S\:-\frac{1}{2^{12}}\right)\\S\:-\frac{S}{2}=\frac{1}{2^3}-\frac{1}{2^{13}}\\S=\frac{1}{2^2}-\frac{1}{2^{12}}\\Substitute\:\:\:in\:\:\:EQ00\:\:we\:\:get\\=>40.\left(\frac{1}{2^2}-\frac{1}{2^{12}}\right)\approx9.990234375.[/tex]
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