What is the sum from n=0 to infinity of 12n(1-n)^2 ?

The sum from n=0 to infinity of 12n(1-n)^2 is diverges

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sum from n=0 to infinity of 12n(1-n)^2

n = 0 \sum \infty 12 n (1 - n)^{2}

diverges

Solution steps

n = 0 \sum \infty 12 n (1 - n)^{2}

Apply the constant multiplication rule:

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

= 12 \cdot n = 0 \sum \infty n (1 - n)^{2}

Apply Series Divergence Test:

diverges

= 12 diverges

= diverges

n = 4 \sum \infty \frac{n}{n ^{2} - 5 n + 6}

n = 1 \sum \infty 5^{- n} n!

n = 0 \sum \infty \frac{n}{9 n ^{2} + 1}

k = 0 \sum \infty (\frac{3}{5})^{k}

n = 1 \sum \infty 5 (\frac{1}{π})^{n}

What is the sum from n=0 to infinity of 12n(1-n)^2 ?
The sum from n=0 to infinity of 12n(1-n)^2 is diverges