What is the sum from n=1 to infinity of (35)/(-6^n) ?

The sum from n=1 to infinity of (35)/(-6^n) is -7

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sum from n=1 to infinity of (35)/(-6^n)

n = 1 \sum \infty \frac{35}{- 6 ^{n}}

- 7

Solution steps

n = 1 \sum \infty \frac{35}{- 6 ^{n}}

Apply the constant multiplication rule:

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

= \frac{35}{- 1} \cdot n = 1 \sum \infty \frac{1}{6 ^{n}}

Simplify

= (- 35) \cdot n = 1 \sum \infty \frac{1}{6 ^{n}}

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

= (- 35) (n = 0 \sum \infty \frac{1}{6 ^{n}} - \frac{1}{6 ^{0}})

n = 0 \sum \infty \frac{1}{6 ^{n}} = \frac{6}{5}

= (- 35) (\frac{6}{5} - \frac{1}{6 ^{0}})

Simplify

(- 35) (\frac{6}{5} - \frac{1}{6 ^{0}}) : - 7

= - 7

\int sin (\frac{θ}{2}) d θ

y^{^{''}} - 9 y = (x^{2} - 5) sin (3 x)

y = 7 x, y = x^{2} - 8, y = 0

\frac{d y}{d x} = 2 x + x y

\frac{d}{d x} ((x^{- 2} + x)^{3} (4 x)^{- 2})

What is the sum from n=1 to infinity of (35)/(-6^n) ?
The sum from n=1 to infinity of (35)/(-6^n) is -7