91Ó°ÊÓ

The binomial theorem is a powerful tool in algebra. It allows us to expand expressions of the form \( (x + y)^n \)into a sum of terms involving binomial coefficients. This means each term in the expansion is a combination of the form: \[ (x + y)^n = \binom{n}{0} x^n y^0 + \binom{n}{1} x^{n-1} y^1 + \binom{n}{2} x^{n-2} y^2 + ... + \binom{n}{k} x^{n-k} y^k + ... + \binom{n}{n} x^0 y^n \].
In other words, you are breaking down a complicated binomial expression into a series of simpler terms. Each of these terms contains a certain number of factors from both \( x \)and \( y \). For example, when we expand \( (x + 2)^8 \), each term will involve different powers of \( x \) and \( 2 \). The power combinations along with the respective multiplicative coefficients form the final expanded expression.

To find the coefficient of a particular term, such as the term involving \( x^{6} \) in the expansion of \( (x + 2)^8 \), we use the binomial theorem. By identifying the right term from the general expansion, we can narrow down to the specific term of interest. The term \( x^{6} \) will fit into the form \ (x + y)^n = \binom{8}{2} x^{6} (2)^2 \.
We need to follow several steps in identifying and calculating the coefficient:

• First, identify \( n \) as 8 and integrate it into the expansion formula.
• Find the term where the power of \( x \) matches 6 (\text{i.e., where \( n - k = 6\text{ and } \ y = 2 \).
• Solve for \( k = 2 \).
• Substitute the values to get the term \( \binom{8}{2} x^{6} (2)^2 \).

After identifying \( k \) and plugging in the values, you compute the binomial coefficient value and simplify it to obtain the coefficient.

A binomial coefficient is a key element in the binomial theorem. It's represented as \binom{n}{k} \ and calculated using the formula: \binom{n}{k} = \frac{n!}{k!(n-k)!} \.
This coefficient is critical because it tells us how many ways we can choose \( k \) items from \( n \)items without regard to the order. In our example, \ (x + 2)^8 \, the binomial coefficient \ \binom{8}{2} \ represents the number of ways to choose 2 powers of \( y \) from 8 terms.
For our specific calculation: \ \binom{8}{2} \ becomes: \ \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \. Multiplying this coefficient by all other terms within the context, including our \( 2^2 \), \ (2)^2 = 4 \ produces the term \ 28 \times 4 x^6 = 112 x^6 \. Therefore, the coefficient for \( x^6 \) is 112.