# Discrete and Combinatorial Mathematics question

Can someone explain to me the area boxed below? I am particularly very interested in the explanation of how the book got (1 + 8)^50 because I have NO IDEA how to figure out that x = 1 and y = 8.....Please help me explain in steps in very simple terms. I am a slow learner!!

32. $\boxed{\sum_{i=0}^{50} {50 \choose i} 8^i = (1 + 8)^{50}} = 9^{50} = [(\pm 3)^2]^{50} = (\pm 3)^{100},$ so $x = \pm 3$.

-

This is known as the binomial theorem. It says that for any $a, b, n$, $$(a + b)^n = \sum_{i=0}^n {n \choose i} a^i b^{n-i}.$$ For example, \begin{align*} (a + b)^1 &= a + b \\ (a + b)^2 &= a^2 + 2ab + b^2 \\ (a + b)^3 &= a^3 + 3ab^2 + 3a^2b + b^3 \end{align*}

In your example, I see a sum from $0$ to $50$ of ${50 \choose i}$ times stuff, which immediately looks to me like binomial theorem. Then I see an $8^i$, which can be the $a^i$ in the formula. For $b^{n-i}$, since there is nothing there, I notice that $1^{n-i} = 1$, so I let $b = 1$. This gives by the binomial theorem that your expression equals $(1 + 8)^{50}$.

-
I am not following you at the part where you say "since there is nothing there, I notice that 1^n-1". Can you explain the logic behind why we use 1 and not any other number? I just don't get how you set it to 1 just because there's nothing there? – Ray May 21 '14 at 6:15
@Ray Notice that ${50 \choose i} 8^i = {50 \choose i} 8^i 1^{50 - i}$. You can always multiply by a power of $1$ without changing the value of an integer. – 6005 May 21 '14 at 6:17