Answer :
To expand the expression
[tex]$$ (2x-1)^6, $$[/tex]
we can use the Binomial Theorem. The Binomial Theorem states that
[tex]$$ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k, $$[/tex]
where [tex]$\binom{n}{k}$[/tex] is the binomial coefficient.
For the expression at hand, assign
[tex]$$ a=2x, \quad b=-1, \quad n=6. $$[/tex]
Therefore, the expansion is
[tex]$$ (2x-1)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k} (-1)^k. $$[/tex]
We now evaluate each term:
1. For [tex]$k=0$[/tex]:
[tex]$$
\binom{6}{0}(2x)^{6}(-1)^0 = 1 \cdot (2x)^6 \cdot 1 = 64x^6.
$$[/tex]
2. For [tex]$k=1$[/tex]:
[tex]$$
\binom{6}{1}(2x)^{5}(-1)^1 = 6 \cdot (2x)^5 \cdot (-1) = 6 \cdot 32x^5 \cdot (-1) = -192x^5.
$$[/tex]
3. For [tex]$k=2$[/tex]:
[tex]$$
\binom{6}{2}(2x)^{4}(-1)^2 = 15 \cdot (2x)^4 \cdot 1 = 15 \cdot 16x^4 = 240x^4.
$$[/tex]
4. For [tex]$k=3$[/tex]:
[tex]$$
\binom{6}{3}(2x)^{3}(-1)^3 = 20 \cdot (2x)^3 \cdot (-1) = 20 \cdot 8x^3 \cdot (-1) = -160x^3.
$$[/tex]
5. For [tex]$k=4$[/tex]:
[tex]$$
\binom{6}{4}(2x)^{2}(-1)^4 = 15 \cdot (2x)^2 \cdot 1 = 15 \cdot 4x^2 = 60x^2.
$$[/tex]
6. For [tex]$k=5$[/tex]:
[tex]$$
\binom{6}{5}(2x)^1(-1)^5 = 6 \cdot (2x) \cdot (-1) = 6 \cdot 2x \cdot (-1) = -12x.
$$[/tex]
7. For [tex]$k=6$[/tex]:
[tex]$$
\binom{6}{6}(2x)^0(-1)^6 = 1 \cdot 1 \cdot 1 = 1.
$$[/tex]
Combining all the terms together, we obtain
[tex]$$ (2x-1)^6 = 64x^6 - 192x^5 + 240x^4 - 160x^3 + 60x^2 - 12x + 1. $$[/tex]
Thus, the correct expansion is
[tex]$$ 64x^6 -192x^5 +240x^4 -160x^3 +60x^2 -12x+1, $$[/tex]
which corresponds to option B.
[tex]$$ (2x-1)^6, $$[/tex]
we can use the Binomial Theorem. The Binomial Theorem states that
[tex]$$ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k, $$[/tex]
where [tex]$\binom{n}{k}$[/tex] is the binomial coefficient.
For the expression at hand, assign
[tex]$$ a=2x, \quad b=-1, \quad n=6. $$[/tex]
Therefore, the expansion is
[tex]$$ (2x-1)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k} (-1)^k. $$[/tex]
We now evaluate each term:
1. For [tex]$k=0$[/tex]:
[tex]$$
\binom{6}{0}(2x)^{6}(-1)^0 = 1 \cdot (2x)^6 \cdot 1 = 64x^6.
$$[/tex]
2. For [tex]$k=1$[/tex]:
[tex]$$
\binom{6}{1}(2x)^{5}(-1)^1 = 6 \cdot (2x)^5 \cdot (-1) = 6 \cdot 32x^5 \cdot (-1) = -192x^5.
$$[/tex]
3. For [tex]$k=2$[/tex]:
[tex]$$
\binom{6}{2}(2x)^{4}(-1)^2 = 15 \cdot (2x)^4 \cdot 1 = 15 \cdot 16x^4 = 240x^4.
$$[/tex]
4. For [tex]$k=3$[/tex]:
[tex]$$
\binom{6}{3}(2x)^{3}(-1)^3 = 20 \cdot (2x)^3 \cdot (-1) = 20 \cdot 8x^3 \cdot (-1) = -160x^3.
$$[/tex]
5. For [tex]$k=4$[/tex]:
[tex]$$
\binom{6}{4}(2x)^{2}(-1)^4 = 15 \cdot (2x)^2 \cdot 1 = 15 \cdot 4x^2 = 60x^2.
$$[/tex]
6. For [tex]$k=5$[/tex]:
[tex]$$
\binom{6}{5}(2x)^1(-1)^5 = 6 \cdot (2x) \cdot (-1) = 6 \cdot 2x \cdot (-1) = -12x.
$$[/tex]
7. For [tex]$k=6$[/tex]:
[tex]$$
\binom{6}{6}(2x)^0(-1)^6 = 1 \cdot 1 \cdot 1 = 1.
$$[/tex]
Combining all the terms together, we obtain
[tex]$$ (2x-1)^6 = 64x^6 - 192x^5 + 240x^4 - 160x^3 + 60x^2 - 12x + 1. $$[/tex]
Thus, the correct expansion is
[tex]$$ 64x^6 -192x^5 +240x^4 -160x^3 +60x^2 -12x+1, $$[/tex]
which corresponds to option B.
