Answer
$6006$
Work Step by Step
$\sum ^{12}_{i=1}i\left( i^{2}-1\right) =\sum ^{12}_{i=1}i^{3}-\sum ^{12}_{i=1}i=\dfrac {1}{4}\times n^2\left( n+1\right) ^{2}-\dfrac {1}{2}\times n\left( n+1\right) =\dfrac {1}{4}\times 12^2\left( 12+1\right) ^{2}-\dfrac {1}{2}\times 12\times \left( 12+1\right) =36\times 13^{2}-6\times 13=6006$
