(3х-1)^7
To expand the expression \((3x - 1)^7\), you can use the Binomial Theorem, which states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In this case, let \(a = 3x\), \(b = -1\), and \(n = 7\).
The expansion is given by:
\[ (3x - 1)^7 = \sum_{k=0}^{7} \binom{7}{k} (3x)^{7-k} (-1)^k \]
Calculating the individual terms gives:
When \(k = 0\): \[ \binom{7}{0} (3x)^7 (-1)^0 = 1 \cdot (3x)^7 \cdot 1 = 2187x^7 \]
When \(k = 1\): \[ \binom{7}{1} (3x)^6 (-1)^1 = 7 \cdot (3x)^6 \cdot (-1) = -7 \cdot 729x^6 = -5103x^6 \]
When \(k = 2\): \[ \binom{7}{2} (3x)^5 (-1)^2 = 21 \cdot (3x)^5 \cdot 1 = 21 \cdot 243x^5 = 5103x^5 \]
When \(k = 3\): \[ \binom{7}{3} (3x)^4 (-1)^3 = 35 \cdot (3x)^4 \cdot (-1) = -35 \cdot 81x^4 = -2835x^4 \]
When \(k = 4\): \[ \binom{7}{4} (3x)^3 (-1)^4 = 35 \cdot (3x)^3 \cdot 1 = 35 \cdot 27x^3 = 945x^3 \]
When \(k = 5\): \[ \binom{7}{5} (3x)^2 (-1)^5 = 21 \cdot (3x)^2 \cdot (-1) = -21 \cdot 9x^2 = -189x^2 \]
When \(k = 6\): \[ \binom{7}{6} (3x)^1 (-1)^6 = 7 \cdot (3x)^1 \cdot 1 = 7 \cdot 3x = 21x \]
When \(k = 7\): \[ \binom{7}{7} (3x)^0 (-1)^7 = 1 \cdot 1 \cdot (-1) = -1 \]
Now, combine all the terms:
\[ (3x - 1)^7 = 2187x^7 - 5103x^6 + 5103x^5 - 2835x^4 + 945x^3 - 189x^2 + 21x - 1 \]
So, the final expanded form is:
\[ (3x - 1)^7 = 2187x^7 - 5103x^6 + 5103x^5 - 2835x^4 + 945x^3 - 189x^2 + 21x - 1 \]
