Question:

Find the 13th term in the expansion of (9x−1/3√x)18,x≠0

Answer:

Answer:

Step 1: Write the expansion of (9x−1/3√x)18,x≠0

(9x−1/3√x)18 = (9x)18 - (1/3√x)18

Step 2: Find the 13th term in the expansion

The 13th term in the expansion of (9x)18 is (9x)13.

The 13th term in the expansion of (1/3√x)18 is (1/3√x)13.

Step 3: Calculate the 13th term in the expansion of (9x−1/3√x)18,x≠0

The 13th term in the expansion of (9x−1/3√x)18,x≠0 is (9x)13 - (1/3√x)13.

Question:

Write the general term in the expansion of (x^2−y)^6

Answer:

(x^2)^6 - 6(x^2)(y)^5 + 15(x^2)(y)^4 - 20(x^2)(y)^3 + 15(y)^4 - 6(y)^5 + (y)^6

General Term: a6x2^6 - 6a5x2y5 + 15a4x2y4 - 20a3x2y3 + 15a2y4 - 6a1y5 + a0y6

Question:

Find the 4th term in the expansion of (x−2y)^12

Answer:

Answer:

Step 1: Rewrite (x−2y)^12 as (x^2 - 4xy + 4y^2)^6

Step 2: Use the binomial theorem to expand (x^2 - 4xy + 4y^2)^6

Step 3: The 4th term in the expansion is 6x^4y^2

Question:

Find the coefficient of a^5b^7 in (a−2b)^12 A : 12356 B : -101356 C : 101376 D : -101376

Answer:

Answer: C

Question:

The coefficients of the (r−1)th,rth and (r+1)th terms in the expansion of (x+1)^n are in the ratio 1 : 3 : 5. Find n and r.

Answer:

Step 1: The coefficient of the (r−1)th term is 1, the coefficient of the rth term is 3, and the coefficient of the (r+1)th term is 5.

Step 2: We can represent this as 1:3:5 = (x+1)^n-r : (x+1)^n : (x+1)^n+r.

Step 3: Dividing both sides by (x+1)^n, we get 1:(x+1) : (x+1)^2 = (x+1)^n-r : (x+1)^n : (x+1)^n+r.

Step 4: Comparing coefficients, we get n-r=0, n=r and n+r=2.

Step 5: Solving the equations, we get n=1 and r=1.

Question:

Find the middle terms in the expansion of (3−x^3/6)^7

Answer:

1. Expand (3−x^3/6)^7 using the binomial theorem: (3−x^3/6)^7 = 3^7 - 73^6x^3/6 + 213^5x^6/36 - 353^4x^9/216 + 353^3x^12/1296 - 213^2x^15/7776 + 73x^18/46656 - x^21/279936

2. Find the middle terms: The middle terms are 73^6x^3/6 and 353^3x^12/1296.

Question:

Write the general term in the expansion of (x^2−yx)^12,x≠0

Answer:

The general term in the expansion of (x^2−yx)^12 is a^12b^n, where a = x^2 and b = -yx.

Question:

Find the middle terms in the expansion of (x/3+9y)^10

Answer:

1. (x/3+9y)^10
2. (x^10/3^10 + 10x^9/3^10 * 9y + 45x^8/3^10 * 9^2y^2 + … + 9^10y^10)
3. (10x^9/3^10 * 9y + 45x^8/3^10 * 9^2y^2 + … + 9^10y^10)
4. (10x^9/3^10 * 9y + 180x^8/3^10 * 9^2y^2 + … + 9^10y^10)

Question:

In the expansion of (1+a)^(m+n) prove that coefficients of a^m and a^n are equal.

Answer:

1. (1+a)^(m+n) = (1+a)^m * (1+a)^n

2. Using the binomial theorem, (1+a)^m = Σ k=0 to m (mCk) (1)^(m-k) (a)^k

3. Similarly, (1+a)^n = Σ k=0 to n (nCk) (1)^(n-k) (a)^k

4. Multiplying (2) and (3), (1+a)^(m+n) = Σ k=0 to m (mCk) (1)^(m-k) (a)^k * Σ k=0 to n (nCk) (1)^(n-k) (a)^k

5. Distributing, (1+a)^(m+n) = Σ k=0 to m Σ k=0 to n (mCk) (nCk) (1)^(m-k+n-k) (a)^(k+k)

6. (1)^(m-k+n-k) = 1

7. (a)^(k+k) = (a)^m if k = m and (a)^(k+k) = (a)^n if k = n

8. Therefore, the coefficients of a^m and a^n are (mCm) (nCn) = (mCn).

Question:

Prove that the coefficient of x^n in the expression of (1+x)^2n is twice the coefficient of x^n in the expression of (1+x)^2n−1.

Answer:

1. Expand (1+x)^2n and (1+x)^2n−1 using the binomial theorem.

(1+x)^2n = Σ(nCk)(1^(2n-k))(x^k)

(1+x)^2n−1 = Σ(nCk)(1^(2n-1-k))(x^k)

2. Compare the coefficients of x^n in the two expressions.

Coefficient of x^n in (1+x)^2n = nCn = 1

Coefficient of x^n in (1+x)^2n−1 = nCn−1 = n/(n-1)

3. Prove that the coefficient of x^n in (1+x)^2n is twice the coefficient of x^n in (1+x)^2n−1.

1 = 2(n/(n-1))

2n = n

n = n

Therefore, the coefficient of x^n in (1+x)^2n is twice the coefficient of x^n in (1+x)^2n−1.

Question:

Find a positive value of m for which the coefficient of x^2 in the expansion (1+x)^m is 6

Answer:

(1+x)^m = a_0 + a_1x + a_2x^2 + a_3x^3 + ….

a_2 = coefficient of x^2

a_2 = (m)(m-1)/2

6 = (m)(m-1)/2

6(2) = m(m-1)

12 = m^2 - m

m^2 - m - 12 = 0

(m-4)(m+3) = 0

m = 4 or m = -3

Since we are looking for a positive value of m,

m = 4

Question:

The coefficient of x5 in the expansion of (x+3)^8 is A : 1542 B : 1512 C : 2512 D : 12 E : 4

Answer:

Answer: B : 1512