Evaluate the summation of negative 2 n minus 3, from n equals 2 to 10..

[tex]\bf \sum\limits_{n=2}^{10}~-2n-3[/tex]

so, let's first change the bounds from 1 to 9, so we'll be dropping the indices by 1, in order to do the variable "n".

however, let's use a variable hmmm say "k", where "k" is the value of "n" beginning at 1, thus k = n - 1.

however, if k = n - 1, then k + 1 = n.  So let's use those fellows then,

[tex]\bf \sum\limits_{n=2}^{10}~-2n-3\qquad \begin{cases} k=n-1\\ k+1=n \end{cases}\implies \sum\limits_{k=1}^{9}~-2\stackrel{n}{(k+1)}-3 \\\\\\ \sum\limits_{k=1}^{9}~-2k-2-3\implies \sum\limits_{k=1}^{9}~-2k-5\implies \sum\limits_{k=1}^{9}~-2k-\sum\limits_{k=1}^{9}~5 \\\\\\ -2\sum\limits_{k=1}^{9}~k-\sum\limits_{k=1}^{9}~5\implies -2\left( \cfrac{9(9+1)}{2} \right)\qquad -\qquad (9\cdot 5) \\\\\\ -2(45)-45\implies -90-45\implies -135[/tex]