Question

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

A.) 9.99
B.) 19.95
C.) 29.98
D.) 39.96

Answer 1

The answer would be B

Answer 2

option A

Explanation:

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

Hope you understand, if you've any doubts comment below ;))