What is the sum from k=0 to infinity of-5(3/2)^{k+2} ?

The sum from k=0 to infinity of-5(3/2)^{k+2} is diverges

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sum from k=0 to infinity of-5(3/2)^{k+2}

k = 0 \sum \infty - 5 (\frac{3}{2})^{k + 2}

diverges

Solution steps

k = 0 \sum \infty - 5 (\frac{3}{2})^{k + 2}

Apply the constant multiplication rule:

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

= - 5 \cdot k = 0 \sum \infty (\frac{3}{2})^{k + 2}

Simplify

(\frac{3}{2})^{k + 2} : (\frac{3}{2})^{k} (\frac{3}{2})^{2}

= - 5 \cdot k = 0 \sum \infty (\frac{3}{2})^{k} (\frac{3}{2})^{2}

Apply the constant multiplication rule:

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

= - 5 (\frac{3}{2})^{2} \cdot k = 0 \sum \infty (\frac{3}{2})^{k}

(\frac{3}{2})^{2} = \frac{9}{4}

= - 5 \cdot \frac{9}{4} \cdot k = 0 \sum \infty (\frac{3}{2})^{k}

Apply Series Geometric Test:

diverges

= - 5 \cdot \frac{9}{4} diverges

= diverges

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What is the sum from k=0 to infinity of-5(3/2)^{k+2} ?
The sum from k=0 to infinity of-5(3/2)^{k+2} is diverges