Consider the following sequence, written in closed form
$$ a_n^2-3a_n $$
$(a) \ $ Write the closed form of the corresponding series , where the lower limit is $ \ 6 \ $ and the upper limit is $ \ 9 \ $.
$ (b) \ $ Write that entire series in open form.
$ (c) \ $ Write the second term of the series .
Answer:
$ (a) \ $
I don't know how to start the answer.
I think , let $ a_{n+1}=a_n^2-3a_n \ $
Since lower limit is $ 6 \ $ and upper limit is $ \ 9 \ $ , we should have
$$ 6=a_0 \leq a_1 \leq a_2 \leq \cdots \leq a_{n+1}=9 \ $$
Now in order to find the closed form of the corresponding series we have to evaluate the series $ \ \sum_{n=0}^\infty (a_n^2-3a_n) \ $
But I am unable to find that.
Help me out
I would take $(a)$ to mean $\displaystyle \sum_{n=6}^9 (a_n^2 -3a_n) = (a_6^2 -3a_6) + (a_7^2-3a_7) + (a_8^2-3a_8) + (a_9^2-3a_9).$