Answer :
Sure! Let's solve the question step by step.
### Part A: Expand [tex]\( (x+1)^8 \)[/tex] using the Binomial Theorem
The Binomial Theorem states that:
[tex]\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} \][/tex]
For our problem:
- [tex]\( n = 8 \)[/tex]
- [tex]\( x = x \)[/tex]
- [tex]\( y = 1 \)[/tex]
So we need to expand [tex]\( (x+1)^8 \)[/tex].
Using the Binomial Theorem, we get:
[tex]\[ (x + 1)^8 = \sum_{k=0}^{8} \binom{8}{k} x^{8-k} \cdot 1^k \][/tex]
Since [tex]\( 1^k = 1 \)[/tex] for any [tex]\( k \)[/tex], the expression simplifies to:
[tex]\[ (x + 1)^8 = \sum_{k=0}^{8} \binom{8}{k} x^{8-k} \][/tex]
Now, let's calculate each binomial coefficient [tex]\( \binom{8}{k} \)[/tex] and the corresponding term in the expansion:
[tex]\[
\begin{align*}
\binom{8}{0} &= 1 & x^{8-0} &= x^8 & \Rightarrow & \quad 1 \cdot x^8 = x^8 \\
\binom{8}{1} &= 8 & x^{8-1} &= x^7 & \Rightarrow & \quad 8 \cdot x^7 = 8x^7 \\
\binom{8}{2} &= 28 & x^{8-2} &= x^6 & \Rightarrow & \quad 28 \cdot x^6 = 28x^6 \\
\binom{8}{3} &= 56 & x^{8-3} &= x^5 & \Rightarrow & \quad 56 \cdot x^5 = 56x^5 \\
\binom{8}{4} &= 70 & x^{8-4} &= x^4 & \Rightarrow & \quad 70 \cdot x^4 = 70x^4 \\
\binom{8}{5} &= 56 & x^{8-5} &= x^3 & \Rightarrow & \quad 56 \cdot x^3 = 56x^3 \\
\binom{8}{6} &= 28 & x^{8-6} &= x^2 & \Rightarrow & \quad 28 \cdot x^2 = 28x^2 \\
\binom{8}{7} &= 8 & x^{8-7} &= x^1 & \Rightarrow & \quad 8 \cdot x = 8x \\
\binom{8}{8} &= 1 & x^{8-8} &= x^0 & \Rightarrow & \quad 1 \cdot 1 = 1 \\
\end{align*}
\][/tex]
Combining all the terms, we get:
[tex]\[ (x + 1)^8 = x^8 + 8x^7 + 28x^6 + 56x^5 + 70x^4 + 56x^3 + 28x^2 + 8x + 1 \][/tex]
### Part B: Methods to identify the correct answer to Part A
Let's look at the possible methods listed:
1. The row corresponding to [tex]\( (a+b)^8 \)[/tex] in Pascal's Triangle has the terms 1, 8, 16, 24, 32, 24, 16, 8, and 1.
- This row is incorrect because the binomial coefficients for [tex]\( (x+1)^8 \)[/tex] should be 1, 8, 28, 56, 70, 56, 28, 8, 1.
2. The row corresponding to [tex]\( (a+b)^8 \)[/tex] in Pascal's Triangle has the terms 1, 7, 21, 35, 35, 21, 7, 1, and 1.
- This row is incorrect because the correct binomial coefficients are, again, 1, 8, 28, 56, 70, 56, 28, 8, 1.
3. The row corresponding to [tex]\( (a+b)^8 \)[/tex] in Pascal's Triangle has the terms 1, 8, 28, 56, 70, 56, 28, 8, and 1.
- This is the correct row for the binomial coefficients for [tex]\( (x+1)^8 \)[/tex].
4. [tex]\[ \sum_{k=0}^8 \binom{8}{k} x^{8-k} 1^k = \sum_{k=0}^8 \binom{8}{k} x^{8-k} \][/tex]
- This is a correct representation using the Binomial Theorem.
Therefore, the correct methods are:
- The row corresponding to [tex]\( (a+b)^8 \)[/tex] in Pascal's Triangle has the terms 1, 8, 28, 56, 70, 56, 28, 8, and 1.
- [tex]\[ \sum_{k=0}^8 \binom{8}{k} x^{8-k} \][/tex]
### Part A: Expand [tex]\( (x+1)^8 \)[/tex] using the Binomial Theorem
The Binomial Theorem states that:
[tex]\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} \][/tex]
For our problem:
- [tex]\( n = 8 \)[/tex]
- [tex]\( x = x \)[/tex]
- [tex]\( y = 1 \)[/tex]
So we need to expand [tex]\( (x+1)^8 \)[/tex].
Using the Binomial Theorem, we get:
[tex]\[ (x + 1)^8 = \sum_{k=0}^{8} \binom{8}{k} x^{8-k} \cdot 1^k \][/tex]
Since [tex]\( 1^k = 1 \)[/tex] for any [tex]\( k \)[/tex], the expression simplifies to:
[tex]\[ (x + 1)^8 = \sum_{k=0}^{8} \binom{8}{k} x^{8-k} \][/tex]
Now, let's calculate each binomial coefficient [tex]\( \binom{8}{k} \)[/tex] and the corresponding term in the expansion:
[tex]\[
\begin{align*}
\binom{8}{0} &= 1 & x^{8-0} &= x^8 & \Rightarrow & \quad 1 \cdot x^8 = x^8 \\
\binom{8}{1} &= 8 & x^{8-1} &= x^7 & \Rightarrow & \quad 8 \cdot x^7 = 8x^7 \\
\binom{8}{2} &= 28 & x^{8-2} &= x^6 & \Rightarrow & \quad 28 \cdot x^6 = 28x^6 \\
\binom{8}{3} &= 56 & x^{8-3} &= x^5 & \Rightarrow & \quad 56 \cdot x^5 = 56x^5 \\
\binom{8}{4} &= 70 & x^{8-4} &= x^4 & \Rightarrow & \quad 70 \cdot x^4 = 70x^4 \\
\binom{8}{5} &= 56 & x^{8-5} &= x^3 & \Rightarrow & \quad 56 \cdot x^3 = 56x^3 \\
\binom{8}{6} &= 28 & x^{8-6} &= x^2 & \Rightarrow & \quad 28 \cdot x^2 = 28x^2 \\
\binom{8}{7} &= 8 & x^{8-7} &= x^1 & \Rightarrow & \quad 8 \cdot x = 8x \\
\binom{8}{8} &= 1 & x^{8-8} &= x^0 & \Rightarrow & \quad 1 \cdot 1 = 1 \\
\end{align*}
\][/tex]
Combining all the terms, we get:
[tex]\[ (x + 1)^8 = x^8 + 8x^7 + 28x^6 + 56x^5 + 70x^4 + 56x^3 + 28x^2 + 8x + 1 \][/tex]
### Part B: Methods to identify the correct answer to Part A
Let's look at the possible methods listed:
1. The row corresponding to [tex]\( (a+b)^8 \)[/tex] in Pascal's Triangle has the terms 1, 8, 16, 24, 32, 24, 16, 8, and 1.
- This row is incorrect because the binomial coefficients for [tex]\( (x+1)^8 \)[/tex] should be 1, 8, 28, 56, 70, 56, 28, 8, 1.
2. The row corresponding to [tex]\( (a+b)^8 \)[/tex] in Pascal's Triangle has the terms 1, 7, 21, 35, 35, 21, 7, 1, and 1.
- This row is incorrect because the correct binomial coefficients are, again, 1, 8, 28, 56, 70, 56, 28, 8, 1.
3. The row corresponding to [tex]\( (a+b)^8 \)[/tex] in Pascal's Triangle has the terms 1, 8, 28, 56, 70, 56, 28, 8, and 1.
- This is the correct row for the binomial coefficients for [tex]\( (x+1)^8 \)[/tex].
4. [tex]\[ \sum_{k=0}^8 \binom{8}{k} x^{8-k} 1^k = \sum_{k=0}^8 \binom{8}{k} x^{8-k} \][/tex]
- This is a correct representation using the Binomial Theorem.
Therefore, the correct methods are:
- The row corresponding to [tex]\( (a+b)^8 \)[/tex] in Pascal's Triangle has the terms 1, 8, 28, 56, 70, 56, 28, 8, and 1.
- [tex]\[ \sum_{k=0}^8 \binom{8}{k} x^{8-k} \][/tex]
