Question 34
Whether $n$ is a positive integer greater than or equal to 5. For numbers $a_{i}$ chosen in the set $\{-1,1\}$, the number $$
S_{n}=a_{1} a_{2} a_{3} a_{4}+\ldots+a_{n} a_{1} a_{2} a_{3}
$$ is calculated that sums the products of each four Terms $a_{i}$ of consecutive indices, including those starting at $a_{n-2}, a_{n-1}$ and $a_{n}$ and ending in $a_{1}, a_{2}$ and $a_{3}$, respectively.
a. Considering $n=8$, let's start with $a_{1}=a_{2}=\ldots=a_{7}=a_{8}=1$. what is the value of $S_{8}$? if we change $a_{4}=1$ to $a_{4}=-1$ how much is the sum $S_{8}$? After the first exchange, we exchange $a_{5}=1$ for $a_{5}=-1$. After this second exchange, how much is $S_{8}$ worth?
b. For each exchange from 1 to -1, how many installments change in value? What are the possible variations in the value of $S_{8}$ when making an exchange?
c. Show that for any eight values of $a_{1}, a_{2}, \ldots, a_{7}$ and $a_{8}$ in the set $\{-1,1\}$ the sum $S_{8}$ always results in a multiple number of 4.
d. for a certain value of $n$ and a certain choice of numbers $a_{i}$ in the set $\{-1,1\}$ The sum $$
S_{n}=a_{1} a_{2} a_{3} a_{4}+\ldots+a_{n} a_{1} a_{2} a_{3}
$$ resulted in zero. Prove that $n$ is necessarily a multiple number of 4.
S_{n}=a_{1} a_{2} a_{3} a_{4}+\ldots+a_{n} a_{1} a_{2} a_{3}
$$ is calculated that sums the products of each four Terms $a_{i}$ of consecutive indices, including those starting at $a_{n-2}, a_{n-1}$ and $a_{n}$ and ending in $a_{1}, a_{2}$ and $a_{3}$, respectively.
a. Considering $n=8$, let's start with $a_{1}=a_{2}=\ldots=a_{7}=a_{8}=1$. what is the value of $S_{8}$? if we change $a_{4}=1$ to $a_{4}=-1$ how much is the sum $S_{8}$? After the first exchange, we exchange $a_{5}=1$ for $a_{5}=-1$. After this second exchange, how much is $S_{8}$ worth?
b. For each exchange from 1 to -1, how many installments change in value? What are the possible variations in the value of $S_{8}$ when making an exchange?
c. Show that for any eight values of $a_{1}, a_{2}, \ldots, a_{7}$ and $a_{8}$ in the set $\{-1,1\}$ the sum $S_{8}$ always results in a multiple number of 4.
d. for a certain value of $n$ and a certain choice of numbers $a_{i}$ in the set $\{-1,1\}$ The sum $$
S_{n}=a_{1} a_{2} a_{3} a_{4}+\ldots+a_{n} a_{1} a_{2} a_{3}
$$ resulted in zero. Prove that $n$ is necessarily a multiple number of 4.