Find the middle term in the expansion of (2x+3y)^12
There will be 13 terms, so the middle term is term #7
Term(7) = C(12,6)(2x)^6 (3y)^6
= 924(64x^6)(729y^6)
= 43110144 x^6 y^6
To find the middle term in the expansion of (2x + 3y)^12, we can use the binomial theorem. According to the binomial theorem, the term in the (k+1)th position of the expansion is given by the formula:
Coefficients * (first term)^m * (second term)^(k - m)
Where:
- Coefficients are the binomial coefficients, which are given by the formula: (n choose k), where n is the power of the binomial (12 in this case) and k is the term number.
- The first term is 2x.
- The second term is 3y.
- m ranges from 0 to k, representing the power of the first term.
To find the middle term, we need to find the term number k for which (k+1) is half of the total number of terms in the expansion. In this case, there will be (12 + 1) = 13 terms in the expansion.
The middle term will be the term number k = (13 - 1) / 2 = 12 / 2 = 6.
Now, let's calculate the middle term:
Term number: k = 6
m = 0 to 6
Coefficient: (12 choose 6) = 924
First term: (2x)^(12-0) = (2x)^12
Second term: (3y)^0 = 1
Middle term = 924 * (2x)^12 * 1 * (3y)^(6-0)
= 924 * (2^12) * (x^12) * (3^6) * (y^6)
Simplifying further, the middle term is:
Middle term = 924 * 4096 * (x^12) * 729 * (y^6)
= 282429536 * (x^12) * (y^6)
Therefore, the middle term in the expansion of (2x + 3y)^12 is 282429536 * (x^12) * (y^6).