Question

Evaluate the summation of 20 times 0.5 to the n minus 1 power, from n equals 3 to 12..

Answer 1

The answer would be B

Answer 2

Answer:

option A

Step-by-step explanation:

$\huge\mathcal{Given}\\=>\sum_{n=3}^{12}20.\left(\frac{1}{2}\right)^{n-1}\\=>20\sum_{n=3}^{12}\left(\frac{1}{2}\right)^{n}\frac{1}{2^{-1}}\\=>40\sum_{n=3}^{12}\left(\frac{1}{2}\right)^{n}==>EQ00\\Let\:\:\:S=\sum_{n=3}^{12}\left(\frac{1}{2}\right)^{n}\\S=\frac{1}{2^3}+\frac{1}{2^4}+.......+\frac{1}{2^{11}}+\frac{1}{2^{12}}=>>EQ01\\S\:-\frac{1}{2^{12}}=\frac{1}{2^3}+\frac{1}{2^4}+.......+\frac{1}{2^{11}}==> EQ02\\From\:\:\:EQ01\:\:\:\\S=\frac{1}{2^3}+\frac{1}{2}\left(\frac{1}{2^3}+\frac{1}{2^4}+.......+\frac{1}{2^{11}}\right)\\From\:\:EQ02\:\:\\S=\frac{1}{2^3}+\frac{1}{2}\left(S\:-\frac{1}{2^{12}}\right)\\S\:-\frac{S}{2}=\frac{1}{2^3}-\frac{1}{2^{13}}\\S=\frac{1}{2^2}-\frac{1}{2^{12}}\\Substitute\:\:\:in\:\:\:EQ00\:\:we\:\:get\\=>40.\left(\frac{1}{2^2}-\frac{1}{2^{12}}\right)\approx9.990234375.$

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