What is the sum from n=1 to infinity}e^{-6n of-e^{-6(n+1)} ?

The sum from n=1 to infinity}e^{-6n of-e^{-6(n+1)} is 1/(e^6)

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sum from n=1 to infinity}e^{-6n of-e^{-6(n+1)}

n = 1 \sum \infty e^{- 6 n} - e^{- 6 (n + 1)}

\frac{1}{e ^{6}}

Solution steps

n = 1 \sum \infty e^{- 6 n} - e^{- 6 (n + 1)}

Apply Telescoping Series Test:

\frac{1}{e ^{6}}

= \frac{1}{e ^{6}}

\int (4 x - 3) 5 d x

\int_{- 2}^{6} 4 x - x^{2} - 12 d x

3 x^{2} e^{x} + e^{x} (x^{3} + 7)

f (x) = x^{2} + x, at x = - 3

y^{^{'}} + y = cos (t)

What is the sum from n=1 to infinity}e^{-6n of-e^{-6(n+1)} ?
The sum from n=1 to infinity}e^{-6n of-e^{-6(n+1)} is 1/(e^6)