Answer :
To find the 7th term in the expansion of [tex]\((3c + 2d)^9\)[/tex], we use the binomial theorem. The theorem tells us that the expansion of [tex]\((a + b)^n\)[/tex] can be expressed as:
[tex]\[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. \][/tex]
Here, [tex]\(a = 3c\)[/tex], [tex]\(b = 2d\)[/tex], and [tex]\(n = 9\)[/tex].
The general term in the expansion is given by:
[tex]\[ T_{k+1} = \binom{n}{k} (3c)^{n-k} (2d)^k. \][/tex]
For the 7th term, we need [tex]\(k = 6\)[/tex] (since terms are numbered from [tex]\(k = 0\)[/tex]).
1. Calculate the Binomial Coefficient:
[tex]\[ \binom{9}{6} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84. \][/tex]
2. Calculate the powers:
- [tex]\((3c)^{9-6} = (3c)^3 = 27c^3\)[/tex],
- [tex]\((2d)^6 = 64d^6\)[/tex].
3. Combine these parts to find the term:
[tex]\[ T_7 = 84 \times 27c^3 \times 64d^6. \][/tex]
Now multiply these numbers:
- [tex]\(84 \times 27 = 2268\)[/tex],
- [tex]\(2268 \times 64 = 145152\)[/tex].
So, the 7th term is:
[tex]\[ 145152c^3d^6. \][/tex]
Therefore, the answer is c. [tex]\(145,152 c^3 d^6\)[/tex].
[tex]\[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. \][/tex]
Here, [tex]\(a = 3c\)[/tex], [tex]\(b = 2d\)[/tex], and [tex]\(n = 9\)[/tex].
The general term in the expansion is given by:
[tex]\[ T_{k+1} = \binom{n}{k} (3c)^{n-k} (2d)^k. \][/tex]
For the 7th term, we need [tex]\(k = 6\)[/tex] (since terms are numbered from [tex]\(k = 0\)[/tex]).
1. Calculate the Binomial Coefficient:
[tex]\[ \binom{9}{6} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84. \][/tex]
2. Calculate the powers:
- [tex]\((3c)^{9-6} = (3c)^3 = 27c^3\)[/tex],
- [tex]\((2d)^6 = 64d^6\)[/tex].
3. Combine these parts to find the term:
[tex]\[ T_7 = 84 \times 27c^3 \times 64d^6. \][/tex]
Now multiply these numbers:
- [tex]\(84 \times 27 = 2268\)[/tex],
- [tex]\(2268 \times 64 = 145152\)[/tex].
So, the 7th term is:
[tex]\[ 145152c^3d^6. \][/tex]
Therefore, the answer is c. [tex]\(145,152 c^3 d^6\)[/tex].
