Question

1 of 51 of 5 Items
Question
Using Pascal’s Triangle, write out the expansion of (a+1)3
.(1 point)

Answer 1

Using Pascal’s Triangle, what is the coefficient of the third term in the expansion of (a+b)6
when the expanded polynomial is written in standard form?(1 point)

Answer 2

Write out the binomial expansion of (a+b)4
using the Binomial Theorem.(1 point)

Answer 3

Using the Binomial Theorem, what would be the coefficient of the sixth term in the binomial expansion of (a+b)9
?(1 point)

Answer 4

To expand \((a + 1)^3\) using Pascal's Triangle, we first identify the coefficients from the third row of Pascal’s Triangle, which corresponds to the expansion of \((x + y)^3\).

The third row of Pascal's Triangle is:
\[
1, 3, 3, 1
\]

Using these coefficients, we can write the expansion as follows:

\[
(a + 1)^3 = \binom{3}{0}a^3(1)^0 + \binom{3}{1}a^2(1)^1 + \binom{3}{2}a^1(1)^2 + \binom{3}{3}a^0(1)^3
\]

Substituting in the binomial coefficients:

\[
(a + 1)^3 = 1 \cdot a^3 + 3 \cdot a^2 \cdot 1 + 3 \cdot a \cdot 1^2 + 1 \cdot 1^3
\]

This simplifies to:

\[
(a + 1)^3 = a^3 + 3a^2 + 3a + 1
\]

Thus, the expansion of \((a + 1)^3\) is:

\[
\boxed{a^3 + 3a^2 + 3a + 1}
\]

Answer 5

To find the coefficient of the third term in the expansion of \((a + b)^6\), we can use Pascal's Triangle to identify the appropriate binomial coefficient.

In the binomial expansion \((a + b)^n\), the \(k\)-th term (where \(k\) starts from 0) is given by:

\[
\binom{n}{k} a^{n-k} b^k
\]

For the expansion of \((a + b)^6\), we are interested in the third term.

Since the term index \(k\) starts from 0, the third term corresponds to \(k = 2\).

We can find the coefficient using the binomial coefficient:

\[
\binom{6}{2}
\]

Now let's look at the 6th row of Pascal's Triangle, which provides us the coefficients for \((a+b)^6\):

\[
1, 6, 15, 20, 15, 6, 1
\]

The coefficients correspond to \( \binom{6}{0}, \binom{6}{1}, \binom{6}{2}, \binom{6}{3}, \binom{6}{4}, \binom{6}{5}, \binom{6}{6} \).

Since we are looking for \(\binom{6}{2}\), the coefficient of the third term is:

\[
\binom{6}{2} = 15
\]

Thus, the coefficient of the third term in the expansion of \((a + b)^6\) is:

\[
\boxed{15}
\]

Answer 6

The Binomial Theorem states that:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

For the case of \((a + b)^4\), we have \(n = 4\). According to the theorem, the expansion can be expressed as:

\[
(a + b)^4 = \sum_{k=0}^{4} \binom{4}{k} a^{4-k} b^k
\]

We will break this down term by term:

- For \(k = 0\):
\[
\binom{4}{0} a^{4} b^{0} = 1 \cdot a^4 \cdot 1 = a^4
\]

- For \(k = 1\):
\[
\binom{4}{1} a^{3} b^{1} = 4 \cdot a^3 \cdot b = 4a^3b
\]

- For \(k = 2\):
\[
\binom{4}{2} a^{2} b^{2} = 6 \cdot a^2 \cdot b^2 = 6a^2b^2
\]

- For \(k = 3\):
\[
\binom{4}{3} a^{1} b^{3} = 4 \cdot a \cdot b^3 = 4ab^3
\]

- For \(k = 4\):
\[
\binom{4}{4} a^{0} b^{4} = 1 \cdot 1 \cdot b^4 = b^4
\]

Now, we can combine all these terms:

\[
(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4
\]

Thus, the binomial expansion of \((a + b)^4\) is:

\[
\boxed{a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4}
\]

Answer 7

To find the coefficient of the sixth term in the binomial expansion of \((a + b)^9\), we can use the Binomial Theorem, which states:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In this expansion, the \(k\)-th term (where \(k\) starts from \(0\)) is given by:

\[
\binom{n}{k} a^{n-k} b^k
\]

Since we want the sixth term, we need to find the term corresponding to \(k = 5\) (because \(k\) starts from 0). Here \(n = 9\), so we want:

\[
\text{Coefficient of } T_{6} = \binom{9}{5} a^{9-5} b^5 = \binom{9}{5} a^4 b^5
\]

Now we compute \(\binom{9}{5}\):

\[
\binom{9}{5} = \binom{9}{4} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!}
\]

Calculating this step by step:

\[
\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126
\]

Thus, the coefficient of the sixth term in the expansion of \((a + b)^9\) is:

\[
\boxed{126}
\]