Binomial Theorem Exercise 2
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)^nr : (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)^nr : (x+1)^n : (x+1)^n+r.
Step 4: Comparing coefficients, we get nr=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:

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

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:
 (x/3+9y)^10
 (x^10/3^10 + 10x^9/3^10 * 9y + 45x^8/3^10 * 9^2y^2 + … + 9^10y^10)
 (10x^9/3^10 * 9y + 45x^8/3^10 * 9^2y^2 + … + 9^10y^10)
 (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+a)^(m+n) = (1+a)^m * (1+a)^n

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

Similarly, (1+a)^n = Σ k=0 to n (nCk) (1)^(nk) (a)^k

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

Distributing, (1+a)^(m+n) = Σ k=0 to m Σ k=0 to n (mCk) (nCk) (1)^(mk+nk) (a)^(k+k)

(1)^(mk+nk) = 1

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

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:
 Expand (1+x)^2n and (1+x)^2n−1 using the binomial theorem.
(1+x)^2n = Σ(nCk)(1^(2nk))(x^k)
(1+x)^2n−1 = Σ(nCk)(1^(2n1k))(x^k)
 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/(n1)
 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/(n1))
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)(m1)/2
6 = (m)(m1)/2
6(2) = m(m1)
12 = m^2  m
m^2  m  12 = 0
(m4)(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