Answer :
To find the coefficient of the third term in the expansion of the binomial [tex]\((3x^2 + 2y^3)^4\)[/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]
For [tex]\((3x^2 + 2y^3)^4\)[/tex], we identify:
[tex]\[ a = 3x^2, \quad b = 2y^3, \quad n = 4 \][/tex]
We need to find the third term of the expansion. In the binomial theorem, terms are indexed from [tex]\(k = 0\)[/tex] to [tex]\(k = n\)[/tex], so the third term corresponds to [tex]\(k = 2\)[/tex].
The general term in the expansion is given by:
[tex]\[ T(k) = \binom{4}{k} (3x^2)^{4-k} (2y^3)^k \][/tex]
Substituting [tex]\(k = 2\)[/tex]:
[tex]\[ T(2) = \binom{4}{2} (3x^2)^{4-2} (2y^3)^2 \][/tex]
Now, we calculate each part of this term:
1. Binomial Coefficient [tex]\(\binom{4}{2}\)[/tex]:
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \][/tex]
2. Power of [tex]\(3x^2\)[/tex]:
[tex]\[ (3x^2)^{4-2} = (3x^2)^2 = 3^2 \cdot (x^2)^2 = 9x^4 \][/tex]
3. Power of [tex]\(2y^3\)[/tex]:
[tex]\[ (2y^3)^2 = 2^2 \cdot (y^3)^2 = 4y^6 \][/tex]
Combining these results:
[tex]\[ T(2) = 6 \times 9x^4 \times 4y^6 \][/tex]
[tex]\[ T(2) = 6 \times 36x^4 y^6 \][/tex]
[tex]\[ T(2) = 216x^4 y^6 \][/tex]
So, the third term in the expansion of [tex]\((3x^2 + 2y^3)^4\)[/tex] is:
[tex]\[ 216x^4 y^6 \][/tex]
Therefore, the coefficient of the third term in the expansion of the binomial [tex]\((3x^2 + 2y^3)^4\)[/tex] is [tex]\(216\)[/tex].
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
For [tex]\((3x^2 + 2y^3)^4\)[/tex], we identify:
[tex]\[ a = 3x^2, \quad b = 2y^3, \quad n = 4 \][/tex]
We need to find the third term of the expansion. In the binomial theorem, terms are indexed from [tex]\(k = 0\)[/tex] to [tex]\(k = n\)[/tex], so the third term corresponds to [tex]\(k = 2\)[/tex].
The general term in the expansion is given by:
[tex]\[ T(k) = \binom{4}{k} (3x^2)^{4-k} (2y^3)^k \][/tex]
Substituting [tex]\(k = 2\)[/tex]:
[tex]\[ T(2) = \binom{4}{2} (3x^2)^{4-2} (2y^3)^2 \][/tex]
Now, we calculate each part of this term:
1. Binomial Coefficient [tex]\(\binom{4}{2}\)[/tex]:
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \][/tex]
2. Power of [tex]\(3x^2\)[/tex]:
[tex]\[ (3x^2)^{4-2} = (3x^2)^2 = 3^2 \cdot (x^2)^2 = 9x^4 \][/tex]
3. Power of [tex]\(2y^3\)[/tex]:
[tex]\[ (2y^3)^2 = 2^2 \cdot (y^3)^2 = 4y^6 \][/tex]
Combining these results:
[tex]\[ T(2) = 6 \times 9x^4 \times 4y^6 \][/tex]
[tex]\[ T(2) = 6 \times 36x^4 y^6 \][/tex]
[tex]\[ T(2) = 216x^4 y^6 \][/tex]
So, the third term in the expansion of [tex]\((3x^2 + 2y^3)^4\)[/tex] is:
[tex]\[ 216x^4 y^6 \][/tex]
Therefore, the coefficient of the third term in the expansion of the binomial [tex]\((3x^2 + 2y^3)^4\)[/tex] is [tex]\(216\)[/tex].
