What is the sum from n=0 to infinity of 7/2 (-3/4)^n ?

The sum from n=0 to infinity of 7/2 (-3/4)^n is 2

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sum from n=0 to infinity of 7/2 (-3/4)^n

n = 0 \sum \infty \frac{7}{2} (- \frac{3}{4})^{n}

2

Solution steps

n = 0 \sum \infty \frac{7}{2} (- \frac{3}{4})^{n}

Apply the constant multiplication rule:

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

= \frac{7}{2} \cdot n = 0 \sum \infty (- \frac{3}{4})^{n}

Apply Series Geometric Test:

\frac{4}{7}

= \frac{7}{2} \cdot \frac{4}{7}

Simplify

\frac{7}{2} \cdot \frac{4}{7} : 2

= 2

n = 1 \sum \infty (0.73)^{n - 1}

n = 2 \sum \infty 1^{l n (n)}

n = 0 \sum \infty \frac{2}{n ^{3} - n}

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

k = 0 \sum \infty (\frac{1}{7})^{k}

What is the sum from n=0 to infinity of 7/2 (-3/4)^n ?
The sum from n=0 to infinity of 7/2 (-3/4)^n is 2