Key to "What is the 7th partial sum of ∑i=1∞6(3)i−1?"

alemanica8h2

Answered question

2023-01-15

What is the 7th partial sum of

\sum_{i = 1}^{\infty} 6 {(3)}^{i - 1}

Jonah Curtis

Beginner2023-01-16Added 10 answers

\sum_{i = 1}^{\infty} 6 {(3)}^{i - 1} = \sum_{i = 1}^{\infty} (2 \cdot 3) {(3)}^{i - 1}

∴ \sum_{i = 1}^{\infty} 6 {(3)}^{i - 1} = \sum_{i = 1}^{\infty} (2) {(3)}^{i}

∴ \sum_{i = 1}^{\infty} 6 {(3)}^{i - 1} = 2 \sum_{i = 1}^{\infty} {(3)}^{i}

∴ \sum_{i = 1}^{\infty} 6 {(3)}^{i - 1} = 2 {3^{1} + 3^{2} + 3^{3} + 3^{4} + \dots

∴ \sum_{i = 1}^{\infty} 6 {(3)}^{i - 1} = 2 {3 + 3 \cdot 3 + 3 \cdot 3^{2} + 3 \cdot 3^{3} + \dots}

And the expression in

{\dots}

is a GP with a=r=3 and so we can use the GP sum formula

S_{n} = a \frac{1 - r^{n}}{(1 - r)}

to get:

\sum_{i = 1}^{\infty} 6 {(3)}^{i - 1} = 2 \cdot 3 \frac{1 - 3^{n}}{1 - 3}

∴ \sum_{i = 1}^{\infty} 6 {(3)}^{i - 1} = 6 \frac{3^{n} - 1}{2}

∴ \sum_{i = 1}^{\infty} 6 {(3)}^{i - 1} = 3 (3^{n} - 1)

Thus, the necessary amount is:

\sum_{i = 1}^{7} 6 {(3)}^{i - 1} = 3 (3^{7} - 1) = 6558

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