Binomial Theorem Miscellaneous Excercise
Question:
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end of the expansion of (4√2+1/4√3)n is √6:1
Answer:
Step 1: Rearrange the given equation to isolate the variable ’n'.
(4√2+1/4√3)n/((4√2+1/4√3)n4) = √6:1
Step 2: Multiply both sides of the equation by (4√2+1/4√3)n4 to get rid of the fraction on the left side.
(4√2+1/4√3)n = (√6:1)(4√2+1/4√3)n4
Step 3: Simplify the right side of the equation.
(4√2+1/4√3)n = (4√6+1/4√3)n4
Step 4: Subtract (4√2+1/4√3)n from both sides of the equation.
0 = (4√6+1/4√3)n4  (4√2+1/4√3)n
Step 5: Simplify the right side of the equation.
0 = (4√64√2+1/4√3)n4
Step 6: Divide both sides of the equation by (4√64√2+1/4√3).
n = 4/(4√64√2+1/4√3)
Question:
Find the coefficient of x^5 in the product (1+2x)^6(1−x)^7 using binomial theorem.
Answer:
Answer: Step 1: Use the binomial theorem to expand (1+2x)^6 (1 + 2x)^6 = 1 + 12x + 54x^2 + 216x^3 + 672x^4 + 1728x^5 + 3024x^6
Step 2: Use the binomial theorem to expand (1−x)^7 (1  x)^7 = 1  7x + 21x^2  35x^3 + 35x^4  21x^5 + 7x^6  x^7
Step 3: Multiply the two expansions: (1 + 12x + 54x^2 + 216x^3 + 672x^4 + 1728x^5 + 3024x^6)(1  7x + 21x^2  35x^3 + 35x^4  21x^5 + 7x^6  x^7)
Step 4: Collect all the terms with x^5 1728x^5
Step 5: The coefficient of x^5 in the product (1+2x)^6(1−x)^7 is 1728.
Question:
Find the expansion (3x^2−2ax+3a^2)^3 using binomial theorem.
Answer:
Answer:
Using the binomial theorem,
(3x^2−2ax+3a^2)^3 =
3^3x^6  3^2(2a)x^5 + 3^2(2a)^2x^4  3(2a)^3x^3 + 3(2a)^4x^2  (2a)^5x + (2a)^6
Question:
Find a, b and n in the expansion of (a+b)^n if the first three terms of the expansion are 729,7290 and 30375, respectively. Insert a number k so that a, k, ,b, n forms an AP.
Answer:
Step 1: Find a, b and n in the expansion of (a+b)^n.
We can use the binomial theorem to calculate the coefficients of the expansion. The first three terms are 729, 7290 and 30375, respectively.
Using the binomial theorem, we can calculate the coefficients of the expansion as follows:
729 = (a+b)^n 7290 = n(a+b)^(n1) 30375 = n(n1)(a+b)^(n2)
Solving for a, b and n, we get:
a = 729/30375 b = 7290/30375 n = 3
Step 2: Insert a number k so that a, k, ,b, n forms an AP.
Since a, b and n form an arithmetic progression, we can insert a number k such that a, k, b, n forms an AP.
The general form of an AP is a + (k  1)d, where d is the common difference.
Substituting the values of a and b, we get:
k = (7290/30375  729/30375)/(3  1) + 1 = 5/9 + 1 = 10/9
Therefore, the number k is 10/9.
Question:
If a and b are distinct integers, prove that a−b is a factor of a^n−b^n, whenever n is a positive integer.
Answer:
Proof:
Let a and b be distinct integers.

By the definition of a factor, we need to show that a^n−b^n is divisible by a−b.

We can rewrite a^n−b^n as (a−b)(a^(n1) + a^(n2)b + … + ab^(n2) + b^(n1)).

Since the terms in the parentheses are all divisible by a−b, then a^n−b^n is also divisible by a−b.

Therefore, a−b is a factor of a^n−b^n, whenever n is a positive integer.
Question:
Expand using Binomial Theorem (1+x/2−2/x)^4,x≠0 and let the sum of coefficients of the terms in the expansion be t. Find 10000t
Answer:
(1 + x/2  2/x)^4
= (1 + x/2)^4  4(1 + x/2)^3(2/x) + 6(1 + x/2)^2(2/x)^2  4(1 + x/2)(2/x)^3 + (2/x)^4
= 1 + 4(x/2) + 6(x/2)^2 + 4(x/2)^3 + (x/2)^4  4(1 + 3(x/2) + 3(x/2)^2 + (x/2)^3)(2/x) + 6(1 + 2(x/2) + (x/2)^2)(2/x)^2  4(1 + (x/2))(2/x)^3 + (2/x)^4
= 1 + 4(x/2) + 6(x/2)^2 + 4(x/2)^3 + (x/2)^4  8/x  24(x/2)/x  24(x/2)^2/x  8(x/2)^3/x + 12/x^2 + 48(x/2)/x^2 + 24(x/2)^2/x^2  8/x^3  8(x/2)/x^3 + 1/x^4
Letting the sum of coefficients of the terms in the expansion be t
t = 1 + 4 + 6 + 4 + 1  8  24  24  8 + 12 + 48 + 24  8  8 + 1
t = 104
10000t = 1040000
Question:
Evaluate (√3−√2)^6+(√3+√2)6 A : 678 B : 950 C : 970 D : 670
Answer:
Step 1: Expand the expression (√3−√2)^6+(√3+√2)^6
Step 2: Simplify 3^6  63^32^1/2 + 63^32^1/2 + 2^6
Step 3: Simplify further 3^6 + 2^6
Step 4: Calculate 3^6 + 2^6 = 970
Answer: C : 970
Question:
Find an approximation of (0.99)^5 using the first three of its expansion. A : 0.95 B : 0.97 C : 0.98 D : 89
Answer:
Answer: C : 0.98
Question:
Find value of (a^2+√(a^2−1))^4+(a^2−√(a^2−1))^4 if a=√5.
Answer:
Solution: Step 1: a = √5
Step 2: (a^2 + √(a^2  1))^4 = (5 + 2√4)^4
Step 3: (a^2  √(a^2  1))^4 = (5  2√4)^4
Step 4: (5 + 2√4)^4 + (5  2√4)^4
Step 5: 625 + 625 + (2^4)(4^2)(√4)^4 + (2^4)(4^2)(√4)^4
Step 6: 1250 + (2^4)(4^2)(4) + (2^4)(4^2)(4)
Step 7: 1250 + 256 + (256)
Step 8: 1250
Question:
Find a if the coefficients of x^2 and x^3 in the expansion of (3+ax)^9 are equal.
Answer:
Step 1: Expand (3+ax)^9 using the Binomial Theorem.
(3+ax)^9 = 3^9 + 93^8ax + 363^7a^2x^2 + 843^6a^3x^3 + 1263^5a^4x^4 + …
Step 2: Set the coefficients of x^2 and x^3 equal to each other.
363^7a^2x^2 = 843^6a^3x^3
Step 3: Divide both sides by x^2.
363^7a^2 = 843^6a^3
Step 4: Divide both sides by 3^6.
123a^2 = 28*a^3
Step 5: Divide both sides by a^2.
123 = 28a
Step 6: Divide both sides by 3.
4 = a