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

Solution:

The given series is

\(\sum_{n=3}^{12}20(0.5)^{n-1}\)

a_n = 20(0.5)(n - 1)

Where n = 3 to 12

a₃ = 20(0.5)(3 - 1) = 20 × (0.5)² = 5

a₄ = 20(0.5)(4 - 1) = 20 × (0.5)³ = 2.5

Common ratio r = 2.5/5 = 0.5

As n = 3 to 12, then number of terms = 10, a₃ = 5 and r = 0.5

Sum of 10 terms S = [5(1 - (0.5)10)]/(1 - 0.5)

By further calculation

S = [5(1 - 0.00098)]/0.5

S = (5 × 0.99902)/0.5

So we get,

S = 4.9951/0.5

S = 9.99

Therefore, the sum of the series is 9.99.

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

Summary:

The summation of 20 times 0.5 to the n minus 1 power, from n equals 3 to 12 is 9.99