The binomial theorem states that \((a + b)^n\) can be expanded as:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For \(n = 4\), the expansion of \((a + b)^4\) is given by:
\[ (a + b)^4 = \sum_{k=0}^{4} \binom{4}{k} a^{4-k} b^k \]
Calculating each term in the expansion:
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For \(k = 0\): \[ \binom{4}{0} a^{4-0} b^0 = 1 \cdot a^4 \cdot 1 = a^4 \]
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For \(k = 1\): \[ \binom{4}{1} a^{4-1} b^1 = 4 \cdot a^3 \cdot b = 4a^3b \]
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For \(k = 2\): \[ \binom{4}{2} a^{4-2} b^2 = 6 \cdot a^2 \cdot b^2 = 6a^2b^2 \]
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For \(k = 3\): \[ \binom{4}{3} a^{4-3} b^3 = 4 \cdot a^1 \cdot b^3 = 4ab^3 \]
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For \(k = 4\): \[ \binom{4}{4} a^{4-4} b^4 = 1 \cdot 1 \cdot b^4 = b^4 \]
Now, combining all the terms, we have:
\[ (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \]
Therefore, the binomial expansion of \((a + b)^4\) is:
\[ \boxed{a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4} \]
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