What is the sum from n=0 to infinity of 5n^2 ?

The sum from n=0 to infinity of 5n^2 is diverges

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sum from n=0 to infinity of 5n^2

n = 0 \sum \infty 5 n^{2}

diverges

Solution steps

n = 0 \sum \infty 5 n^{2}

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty n^{2}

Apply Series Divergence Test:

diverges

= 5 diverges

= diverges

n = 2 \sum \infty \frac{1}{3 n + 4}

n = 1 \sum \infty 3 (0.4)^{n - 1}

n = 3 \sum \infty \frac{1}{5 ^{n}}

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

n = 1 \sum \infty \frac{8}{- 3 ^{n}}

What is the sum from n=0 to infinity of 5n^2 ?
The sum from n=0 to infinity of 5n^2 is diverges