What is the result of evaluating this sequence challenge?

[anemone]

A sequence of integers ${x_i}$ is defined as follows:

$x_i=i$ for all $1<i<5$ and

$x_i=(x_1x_2\cdots x_{i-1})-1$ for $i>5$.

Evaluate $\displaystyle x_1x_2\cdots x_{2011}-\sum_{i=1}^{2011} (x_i)^2$.

 

[anemone]

A sequence of integers ${x_i}$ is defined as follows:

$x_i=i$ for all $1<i<5$ and

$x_i=(x_1x_2\cdots x_{i-1})-1$ for $i>5$.

Evaluate $\displaystyle x_1x_2\cdots x_{2011}-\sum_{i=1}^{2011} (x_i)^2$.

Hi MHB,

I want to apologize for not checking the validity of the interval for this sequence because the first function should be defined at $1\le i \le 5$. But I wouldn't have noticed it if Euge didn't let me know of it. Therefore, I owe Euge a thank, and perhaps a cup of coffee as well?

The problem should read:

$x_i=i$ for all $1\le i \le5$ and

$x_i=(x_1x_2\cdots x_{i-1})-1$ for $i>5$.

Evaluate $\displaystyle x_1x_2\cdots x_{2011}-\sum_{i=1}^{2011} (x_i)^2$.

 

[Euge]

Hi MHB,

I want to apologize for not checking the validity of the interval for this sequence because the first function should be defined at $1\le i \le 5$. But I wouldn't have noticed it if Euge didn't let me know of it. Therefore, I owe Euge a thank, and perhaps a cup of coffee as well?

The problem should read:

$x_i=i$ for all $1\le i \le5$ and

$x_i=(x_1x_2\cdots x_{i-1})-1$ for $i>5$.

Evaluate $\displaystyle x_1x_2\cdots x_{2011}-\sum_{i=1}^{2011} (x_i)^2$.

Ok, here is my solution.

Spoiler

We have

$\displaystyle x_1 \cdots x_{2011} -\sum_{i = 1}^{2011} (x_i)^2$

$\displaystyle = x_1 \cdots x_{2011} - \sum_{i = 1}^{5} i^2 - \sum_{i = 6}^{2011} [(x_i - 1)(x_i + 1) + 1]$

$\displaystyle = x_1 \cdots x_{2011} - 55 -\sum_{i = 6}^{2011} (x_i - 1)x_1 \cdots x_{i - 1} - 2006$

$\displaystyle = x_1 \cdots x_{2011} - \sum_{i = 6}^{2011} (x_1 \cdots x_i - x_1 \cdots x_{i - 1}) - 2061$

$\displaystyle = x_1 \cdots x_{2011} - x_1 \cdots x_{2011} + 5! -2061$

$\displaystyle = -1941$.

 

[anemone]

Hey Euge!:)

Thanks for participating and your answer is correct! Well done!(Yes)