What is the sum from n=1 to infinity of (12)/(-5^n) ?

The sum from n=1 to infinity of (12)/(-5^n) is -3

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sum from n=1 to infinity of (12)/(-5^n)

n = 1 \sum \infty \frac{12}{- 5 ^{n}}

- 3

Solution steps

n = 1 \sum \infty \frac{12}{- 5 ^{n}}

Apply the constant multiplication rule:

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

= \frac{12}{- 1} \cdot n = 1 \sum \infty \frac{1}{5 ^{n}}

Simplify

= (- 12) \cdot n = 1 \sum \infty \frac{1}{5 ^{n}}

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

= (- 12) (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}})

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= (- 12) (\frac{5}{4} - \frac{1}{5 ^{0}})

Simplify

(- 12) (\frac{5}{4} - \frac{1}{5 ^{0}}) : - 3

= - 3

w e^{7 w}

f (x) = x^{- 15}

y^{^{'}} + \frac{1}{200 + 3 t} y = 12

\int \frac{1}{9 v ^{2} - 16} d v

3 x^{2} - 2

What is the sum from n=1 to infinity of (12)/(-5^n) ?
The sum from n=1 to infinity of (12)/(-5^n) is -3