What is the sum from k=1 to infinity of 3^k7 ?

The sum from k=1 to infinity of 3^k7 is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=1 to infinity of 3^k7

k = 1 \sum \infty 3^{k} 7

diverges

Solution steps

k = 1 \sum \infty 3^{k} \cdot 7

Apply the constant multiplication rule:

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

= 7 \cdot k = 1 \sum \infty 3^{k}

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

= 7 (k = 0 \sum \infty 3^{k} - 3^{0})

k = 0 \sum \infty 3^{k} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d y}{d x} = 4 y + x^{2} + 7

\frac{\partial}{\partial y} (x^{2} + y^{2})

\int 6 x - 8 d x

\int 7 ln (x^{2} - x + 6) d x

\frac{\partial}{\partial y} (3 x^{2} y^{2} - 1)

What is the sum from k=1 to infinity of 3^k7 ?
The sum from k=1 to infinity of 3^k7 is diverges