What is the sum from n=1 to infinity of 2(3/4)^n ?

The sum from n=1 to infinity of 2(3/4)^n is 6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity of 2(3/4)^n

n = 1 \sum \infty 2 (\frac{3}{4})^{n}

6

Solution steps

n = 1 \sum \infty 2 (\frac{3}{4})^{n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty (\frac{3}{4})^{n}

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

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

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

= 2 (4 - (\frac{3}{4})^{0})

Simplify

= 6

\frac{sec ^{2} ( x )}{4}, 4 cos^{2} (x)

x \to - 2 + lim (\frac{( x + 3 )}{x + 2})

\int \frac{1}{x ^{2} 4 x ^{2} - 36} d x

\int (x + 5) 5 - x d x

f (x) = 8 x sin (x)

What is the sum from n=1 to infinity of 2(3/4)^n ?
The sum from n=1 to infinity of 2(3/4)^n is 6