What is the sum from n=1 to infinity of 8(0.2)^n ?

The sum from n=1 to infinity of 8(0.2)^n is 2

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sum from n=1 to infinity of 8(0.2)^n

n = 1 \sum \infty 8 (0.2)^{n}

2

Solution steps

n = 1 \sum \infty 8 \cdot 0. 2^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 8 \cdot n = 1 \sum \infty 0. 2^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 8 (n = 0 \sum \infty 0. 2^{n} - 0. 2^{0})

n = 0 \sum \infty 0. 2^{n} = 1.25

= 8 (1.25 - 0. 2^{0})

Simplify

= 2

n = 1 \sum \infty (1 + \frac{3}{n})^{n}

x \to - 1 lim (\frac{x ^{3} - 1}{x - 1})

y = 5 x^{3} - 5 x, (1, 0)

n = 1 \sum \infty \frac{2 ^{n} x ^{n}}{n}

\int \frac{x ^{2}}{1 + x ^{2}} d x

What is the sum from n=1 to infinity of 8(0.2)^n ?
The sum from n=1 to infinity of 8(0.2)^n is 2