Question
write out the binomial expansion of (a+b)^4 using binonial theorem
Answers
GPT-4o mini
The binomial theorem states that for any positive integer \( n \), the expansion of \( (a + b)^n \) can be expressed as:
\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]
where \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
For \( n = 4 \), the expression \( (a + b)^4 \) expands as follows:
\[
(a + b)^4 = \sum_{k=0}^{4} \binom{4}{k} a^{4-k} b^k
\]
Calculating each term:
- For \( k = 0 \):
\[
\binom{4}{0} a^{4-0} b^0 = 1 \cdot a^4 \cdot 1 = a^4
\]
- For \( k = 1 \):
\[
\binom{4}{1} a^{4-1} b^1 = 4 \cdot a^3 \cdot b = 4a^3b
\]
- For \( k = 2 \):
\[
\binom{4}{2} a^{4-2} b^2 = 6 \cdot a^2 \cdot b^2 = 6a^2b^2
\]
- For \( k = 3 \):
\[
\binom{4}{3} a^{4-3} b^3 = 4 \cdot a^1 \cdot b^3 = 4ab^3
\]
- For \( k = 4 \):
\[
\binom{4}{4} a^{4-4} b^4 = 1 \cdot 1 \cdot b^4 = b^4
\]
Combining all these terms, the complete expansion of \( (a+b)^4 \) is:
\[
(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4
\]
\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]
where \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
For \( n = 4 \), the expression \( (a + b)^4 \) expands as follows:
\[
(a + b)^4 = \sum_{k=0}^{4} \binom{4}{k} a^{4-k} b^k
\]
Calculating each term:
- For \( k = 0 \):
\[
\binom{4}{0} a^{4-0} b^0 = 1 \cdot a^4 \cdot 1 = a^4
\]
- For \( k = 1 \):
\[
\binom{4}{1} a^{4-1} b^1 = 4 \cdot a^3 \cdot b = 4a^3b
\]
- For \( k = 2 \):
\[
\binom{4}{2} a^{4-2} b^2 = 6 \cdot a^2 \cdot b^2 = 6a^2b^2
\]
- For \( k = 3 \):
\[
\binom{4}{3} a^{4-3} b^3 = 4 \cdot a^1 \cdot b^3 = 4ab^3
\]
- For \( k = 4 \):
\[
\binom{4}{4} a^{4-4} b^4 = 1 \cdot 1 \cdot b^4 = b^4
\]
Combining all these terms, the complete expansion of \( (a+b)^4 \) is:
\[
(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4
\]