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A sequence of integers is defined as follows:  $$a_i = i$$ for $$1$$$$\le i \le 5,$$ and $$a_i = a_1 a_2 \dotsm a_{i - 1} - 1$$ for $$i > 5$$

Evaluate $$a_1 a_2 \dotsm a_{2011} - \sum_{i = 1}^{2011} a_i^2$$

So for this question, the first step I took was recognizing that $$a_1 a_2 \dotsm a_{2011} = a_{2012}+1$$

Additionally, for i > 5,  $$a_{n+1} = {a_n}^2+a_n-1$$ through some algebraic manipulation.

Moving terms of the equation to one side, it follows that $$a_{n+1} -a_n+1= {a_n}^2$$

Thus, $$a_1 a_2 \dotsm a_{2011} - \sum_{i = 1}^{2011} a_i^2$$ would be equal to $$a_1 a_2 \dotsm a_{2011} - \sum_{i = 1}^{2011} a_{n+1}-a_n+1$$.  From this, we have a telescoping series for all terms after a_6, as $$a_1, a_2, a_3, a_4, a_5$$ do not follow this pattern.

After writing it all out, we find that the sequence $$\sum_{i = 1}^{2011} a_{n+1}-a_n+1$$ simplifies to $$a_1^2+a_2^2+a_3^2+a_4^2+a_5^2$$+$$a_{2012}-a_6+2006$$

This is equal to $$55-(5!-1)+a_{2012}+2006$$.  Plugging this into the original sequence, we have it equal to

$$a_{2012}+1-(55-(5!-1)+a_{2012}+2006)$$, simplifying to

$$1-55+(5!-1)-2006$$, or $$719-2060$$ or $$-1341$$.

I am very grateful for any responses that I get, this is my last attempt at this question, so I really hope I can get it right (if I do not, I have to do many more questions to compensate lol).

Thank you so much!

Jun 9, 2021

#1
0

Everything looks right to me.  Good job!

Jun 9, 2021
#2
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F

It turns out, 5!-1 is not 719, but in fact 119.

Jun 14, 2021