What is the sum from n=3 to infinity of 1/2*(5/7)^n ?

The sum from n=3 to infinity of 1/2*(5/7)^n is 125/196

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sum from n=3 to infinity of 1/2*(5/7)^n

n = 3 \sum \infty \frac{1}{2} \cdot (\frac{5}{7})^{n}

\frac{125}{196}

Solution steps

n = 3 \sum \infty \frac{1}{2} (\frac{5}{7})^{n}

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 3 \sum \infty (\frac{5}{7})^{n}

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

= \frac{1}{2} (n = 0 \sum \infty (\frac{5}{7})^{n} - (\frac{5}{7})^{0} - (\frac{5}{7})^{1} - (\frac{5}{7})^{2})

n = 0 \sum \infty (\frac{5}{7})^{n} = \frac{7}{2}

= \frac{1}{2} (\frac{7}{2} - (\frac{5}{7})^{0} - (\frac{5}{7})^{1} - (\frac{5}{7})^{2})

Simplify

\frac{1}{2} (\frac{7}{2} - (\frac{5}{7})^{0} - (\frac{5}{7})^{1} - (\frac{5}{7})^{2}) : \frac{125}{196}

= \frac{125}{196}

\int sec^{2} (x) sin (x) d x

\int x^{0.5} d x

\int \frac{1}{x ^{3} + 4 x} d x

f (x) = 4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 2 x

f (x) = \frac{2}{x}, at \frac{1}{16}

What is the sum from n=3 to infinity of 1/2*(5/7)^n ?
The sum from n=3 to infinity of 1/2*(5/7)^n is 125/196