Given ∑p=1n(−p−1)=−90\displaystyle {\sum_{{{p}={1}}}^{{n}}}{\left(-{p}-{1}\right)}=-{90}

Determine:

Question 1

...the type of series.

(1)

Question 2

... =\displaystyle =

(2)

Question 3



...the value(s) of n.

Formula

    n=\displaystyle _{n}=  

Stocktake

n=\displaystyle {n}=

    n\displaystyle _{n} =\displaystyle =

    a\displaystyle {a} =\displaystyle =

    =\displaystyle =

Substitute

    =\displaystyle =  

Simplify

      (in the form an2+bn+c=0\displaystyle {a}{n}^{{2}}+{b}{n}+{c}={0})

Solve

   n=\displaystyle {n}= (not applicable) or n=\displaystyle {n}=

(5)

[8]

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