Sum

Find the sum of the Geometric series 3 + 6 + 12 + …….. + 1536

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#### Solution

3 + 6 + 12 + …. + 1536

a = 3, r = `6/3` = 2

t_{n} = 1536

a.r^{n–1} = 1536 ⇒ 3(2^{n–1}) = 1536

2^{n–1} = `1536/3` ⇒ 2^{n–1} = 512

2^{n–1} = 2^{9}

∴ n – 1 = 9

n = 9 + 1 = 10

Number of terms = 10

S_{n} = `("a"("r"^"n" - 1))/("r" - 1)`

S_{n} = `(3(2^10 - 1))/(2 - 1)`

= `(3(1024 - 1))/1`

= 3 × 1023

= 3069

∴ Sum of the series is 3069

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