What is the sum from n=6 to infinity of n^2-16 ?

The sum from n=6 to infinity of n^2-16 is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=6 to infinity of n^2-16

n = 6 \sum \infty n^{2} - 16

diverges

Solution steps

n = 6 \sum \infty n^{2} - 16

Multiply by the conjugate of

n^{2} - 16 : \frac{n ^{4} - 256}{n ^{2} + 16}

= n = 6 \sum \infty \frac{n ^{4} - 256}{n ^{2} + 16}

Apply Series Divergence Test:

diverges

= diverges

2 x + y = 4, y = x, y = 0

\int 7 e^{4 x + e^{4 x}} d x

\frac{d}{d s} (\frac{( s - 3 ) ( s - 5 )}{( s + 1 ) ( s + 2 )})

\frac{d}{d t} (3 - 2 t)

\frac{1 - x ^{2} + x ^{4}}{x ^{2} \cdot x}

What is the sum from n=6 to infinity of n^2-16 ?
The sum from n=6 to infinity of n^2-16 is diverges