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Calculate the sum of  NINE terms (sub in n=9).....  Then Calculate the sum of TEN terms (sub in n=10)...

   the 10th term is the difference  

The sum of the first n terms of a certain sequence is

$\huge{n(n+1)(n+2)}$
Find the tenth term of the sequence.

$\mathbf{a_n = \ ?}$

$\begin{array}{|rcll|} \hline a_n &=& n(n+1)(n+2) - (n-1)\Big((n-1)+1)((n-1)+2 \Big) \\ a_n &=& n(n+1)(n+2) - (n-1)n(n+1) \\ a_n &=& n(n+1)\Big((n+2)-(n-1)\Big) \\ a_n &=& n(n+1)( n+2 - n+1) \\ a_n &=& 3n(n+1) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a_{10} &=& 3\cdot 10 \cdot (10+1) \\ a_{10} &=& 33\cdot 10 \\ \mathbf{a_{10}} & \mathbf{=} & \mathbf{330} \\ \hline \end{array}$