What is expand e^{-n+1}(1+1/(n+1)) ?

The solution to expand e^{-n+1}(1+1/(n+1)) is e^{-n+1}+(e^{-n+1})/(n+1)

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expand e^{-n+1}(1+1/(n+1))

expand

e^{- n + 1} (1 + \frac{1}{n + 1})

e^{- n + 1} + \frac{e ^{- n + 1}}{n + 1}

Solution steps

e^{- n + 1} (1 + \frac{1}{n + 1})

Apply the distributive law:

a (b + c) = ab + a c

a = e^{- n + 1}, b = 1, c = \frac{1}{n + 1}

= e^{- n + 1} \cdot 1 + e^{- n + 1} \frac{1}{n + 1}

= 1 \cdot e^{- n + 1} + e^{- n + 1} \frac{1}{n + 1}

Simplify

1 \cdot e^{- n + 1} + e^{- n + 1} \frac{1}{n + 1} : e^{- n + 1} + \frac{e ^{- n + 1}}{n + 1}

= e^{- n + 1} + \frac{e ^{- n + 1}}{n + 1}

- cos (x) sec (x) tan (x) + 2 sec (x) sin (x)

a (2 - 4) b (- 3.2)

\frac{4}{5} \times \frac{7}{9} \times \frac{1}{2}

\frac{x ^{2} + 2 x - 35}{5 x ( x ^{2} - 1 )}

[(\frac{1}{2})^{2} \cdot (\frac{3}{5})^{2}]^{2}

What is expand e^{-n+1}(1+1/(n+1)) ?
The solution to expand e^{-n+1}(1+1/(n+1)) is e^{-n+1}+(e^{-n+1})/(n+1)