General Term and Middle Term

[MathDefinition title="General Term"] The general term (or (r+1)th term) in the expansion of (a + b)^n is: T_{r+1} = nCr * a^{n-r} * b^r, where r = 0, 1, 2, ..., n [/MathDefinition]

[MathDefinition title="Middle Term"] In the expansion of (a + b)^n:

• If n is even: Middle term = T_{n/2 + 1}
• If n is odd: Middle terms = T_{(n+1)/2} and T_{(n+3)/2} [/MathDefinition]

[MathKeyFormula title="Key Formulas"]

1. General term: T_{r+1} = nCr * a^{n-r} * b^r
2. For (1 + x)^n: T_{r+1} = nCr * x^r
3. For (a + bx)^n: T_{r+1} = nCr * a^{n-r} * (bx)^r
4. Middle term (n even): T_{n/2 + 1}
5. Middle term (n odd): T_{(n+1)/2} and T_{(n+3)/2} [/MathKeyFormula]

[MathNote title="Finding Specific Terms"]

1. Identify a, b, and n from the expansion
2. Use the general term formula T_{r+1} = nCr * a^{n-r} * b^r
3. Set the power of x to the desired value
4. Solve for r
5. Calculate the term using the value of r [/MathNote]

[MathExample title="Example 1: Finding a Specific Term"] Problem: Find the 5th term in the expansion of (3x - 2y)^7.

Solution: Here a = 3x, b = -2y, n = 7

General term: T_{r+1} = 7Cr * (3x)^{7-r} * (-2y)^r

For the 5th term, r = 4: T5 = 7C4 * (3x)^3 * (-2y)^4

7C4 = 35 (3x)^3 = 27x^3 (-2y)^4 = 16y^4

T5 = 35 * 27x^3 * 16y^4 = 15120x^3y^4

Therefore, the 5th term is 15120x^3y^4. [/MathExample]

[MathExample title="Example 2: Middle Term Even n"] Problem: Find the middle term in the expansion of (x + 1/x)^8.

Solution: Here n = 8 (even), so the middle term is T_{8/2 + 1} = T5

General term: T_{r+1} = 8Cr * x^{8-r} * (1/x)^r = 8Cr * x^{8-2r}

For T5, r = 4: T5 = 8C4 * x^{8-8} = 8C4 * x^0 = 8C4

8C4 = 8!/(4! * 4!) = 70

Therefore, the middle term is 70. [/MathExample]

[MathExample title="Example 3: Middle Term Odd n"] Problem: Find the middle terms in the expansion of (x/3 + 9y)^5.

Solution: Here n = 5 (odd), so middle terms are T3 and T4

General term: T_{r+1} = 5Cr * (x/3)^{5-r} * (9y)^r

For T3 (r = 2): T3 = 5C2 * (x/3)^3 * (9y)^2 = 10 * (x^3/27) * 81y^2 = 30x^3y^2

For T4 (r = 3): T4 = 5C3 * (x/3)^2 * (9y)^3 = 10 * (x^2/9) * 729y^3 = 810x^2y^3

The middle terms are 30x^3y^2 and 810x^2y^3. [/MathExample]

[MathExample title="Example 4: Term Independent of x"] Problem: Find the term independent of x in the expansion of (x + 1/x^2)^9.

Solution: General term: T_{r+1} = 9Cr * x^{9-r} * (1/x^2)^r = 9Cr * x^{9-3r}

For term independent of x: 9 - 3r = 0, so r = 3

T4 = 9C3 = 9!/(3! * 6!) = 84

Therefore, the term independent of x is 84. [/MathExample]