The equation shows the beginning of a problem using the Binomial Theorem to expand a power of a binomial.
(a+b)3= _[blank]_
Which equation correctly fills in the blank to complete the Binomial Theorem?
∑3k=0(3k)a3−kbk=1a3+3a2b+3ab2+1b3
∑3k=1(3k)a3bk=3a3+1a2b+1ab2+3b3
∑3k=0a3−kbk=1a3+2ab2+1b3
∑3k=0(k3)a3bk=1k3+3k2b+3kb2+1b3
(3k)a3−kbk=1a3+4a2b+6ab2+4b3
Better take another look at Pascal's Triangle.
for (a+b)^3 the coefficients are 1,3,3,1 not 1,4,6,4
The rows always start and end with 1.
and your notation is very hard to parse.
The equation that correctly fills in the blank to complete the Binomial Theorem is:
∑3k=0(3k)a3−kbk=1a3+3a2b+3ab2+1b3
To fill in the blank correctly, we need to use the Binomial Theorem, which states that for any positive integer n:
(a + b)^n = ∑(nCk)(a^(n-k))(b^k)
Where:
- ∑ represents the summation notation, with k starting from 0 to n,
- nCk represents the binomial coefficient, also known as "n choose k," which is equal to n! / (k!(n-k)!),
- a^(n-k) represents a raised to the power of (n-k),
- b^k represents b raised to the power of k.
Now, let's look at the given options:
Option 1: ∑3k=0(3k)a3−kbk=1a3+3a2b+3ab2+1b3.
This option correctly uses the summation notation and the correct binomial coefficients. It also uses the correct exponents for both a and b. Therefore, this is a valid equation.
Option 2: ∑3k=1(3k)a3bk=3a3+1a2b+1ab2+3b3.
This option starts the summation from k=1 instead of k=0. According to the Binomial Theorem, we need to include the term where k=0. Therefore, this is not a valid equation.
Option 3: ∑3k=0a3−kbk=1a3+2ab2+1b3.
This option is missing the binomial coefficients (nCk), which are necessary for the Binomial Theorem. Therefore, this is not a valid equation.
Option 4: ∑3k=0(k3)a3bk=1k3+3k2b+3kb2+1b3.
This option uses (k^3) instead of (3^k) as the binomial coefficients. The binomial coefficients should represent (3Ck), not (k^3). Therefore, this is not a valid equation.
Option 5: (3k)a3−kbk=1a3+4a2b+6ab2+4b3.
This option does not use the summation notation (∑) or include the binomial coefficients (nCk). Therefore, this is not a valid equation.
Therefore, the correct equation that fills in the blank is:
∑3k=0(3k)a3−kbk=1a3+3a2b+3ab2+1b3.