Answer :
Certainly! Let's walk through solving [tex]\( p = \binom{-6}{1} \)[/tex] and [tex]\( q = \binom{13}{14} \)[/tex] first, and then we will calculate [tex]\( 4p + q \)[/tex].
### Step 1: Calculating [tex]\( p = \binom{-6}{1} \)[/tex]
The binomial coefficient [tex]\(\binom{n}{k}\)[/tex] is typically defined for non-negative integers [tex]\(n\)[/tex] and [tex]\(k\)[/tex] such that [tex]\(0 \leq k \leq n\)[/tex]. However, it can also be extended to include cases where [tex]\(n\)[/tex] can be a negative integer using the general formula for binomial coefficients:
[tex]\[ \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \][/tex]
For [tex]\( n = -6 \)[/tex] and [tex]\( k = 1 \)[/tex]:
[tex]\[ \binom{-6}{1} = \frac{-6}{1!} = -6 \][/tex]
So, [tex]\( p = -6 \)[/tex].
### Step 2: Calculating [tex]\( q = \binom{13}{14} \)[/tex]
For any binomial coefficient [tex]\(\binom{n}{k}\)[/tex] where [tex]\( k > n \)[/tex], the value is 0 because there are no ways to choose more elements than are available:
[tex]\[ \binom{13}{14} = 0 \][/tex]
So, [tex]\( q = 0 \)[/tex].
### Step 3: Calculating [tex]\( 4p + q \)[/tex]
Now we combine the results:
[tex]\[ 4p + q = 4(-6) + 0 = -24 + 0 = -24 \][/tex]
### Step 4: Writing the answer as a column vector
The final answer as a column vector:
[tex]\[ \begin{pmatrix} -24 \end{pmatrix} \][/tex]
Thus, [tex]\( 4p + q = \begin{pmatrix} -24 \end{pmatrix} \)[/tex].
### Step 1: Calculating [tex]\( p = \binom{-6}{1} \)[/tex]
The binomial coefficient [tex]\(\binom{n}{k}\)[/tex] is typically defined for non-negative integers [tex]\(n\)[/tex] and [tex]\(k\)[/tex] such that [tex]\(0 \leq k \leq n\)[/tex]. However, it can also be extended to include cases where [tex]\(n\)[/tex] can be a negative integer using the general formula for binomial coefficients:
[tex]\[ \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \][/tex]
For [tex]\( n = -6 \)[/tex] and [tex]\( k = 1 \)[/tex]:
[tex]\[ \binom{-6}{1} = \frac{-6}{1!} = -6 \][/tex]
So, [tex]\( p = -6 \)[/tex].
### Step 2: Calculating [tex]\( q = \binom{13}{14} \)[/tex]
For any binomial coefficient [tex]\(\binom{n}{k}\)[/tex] where [tex]\( k > n \)[/tex], the value is 0 because there are no ways to choose more elements than are available:
[tex]\[ \binom{13}{14} = 0 \][/tex]
So, [tex]\( q = 0 \)[/tex].
### Step 3: Calculating [tex]\( 4p + q \)[/tex]
Now we combine the results:
[tex]\[ 4p + q = 4(-6) + 0 = -24 + 0 = -24 \][/tex]
### Step 4: Writing the answer as a column vector
The final answer as a column vector:
[tex]\[ \begin{pmatrix} -24 \end{pmatrix} \][/tex]
Thus, [tex]\( 4p + q = \begin{pmatrix} -24 \end{pmatrix} \)[/tex].
