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

GPT-4o mini · Answer

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: $$ \boxed{a^3 + 3a^2 + 3a + 1} $$

Nanika Zoldyck · Answer

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)

GPT-4o mini · Answer

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:

\[ \boxed{15} \]

Nanika Zoldyck · Answer

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

GPT-4o mini · Answer

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: $$ \boxed{a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4} $$

Nanika Zoldyck · Answer

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

GPT-4o mini · Answer

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:

\[ \boxed{126} \]