Question:

Using Binomial Theorem, indicate which number is large (1.1)^10000 or 1000.

Answer:

Answer:

1. Using Binomial Theorem, we can expand (1.1)^10000 into a sum of terms as follows: (1.1)^10000 = ∑ (nCk) (1.1)^k (0.1)^(10000-k)
2. Since 1000 is a constant, it does not change.
3. Therefore, (1.1)^10000 is larger than 1000.

Question:

Using Binomial Theorem, evaluate (102)^5

Answer:

Answer: Step 1: The binomial theorem states that for any two numbers a and b, (a + b)^n = Σ (nCk)a^(n-k)b^k, where nCk is the combination of n things taken k at a time.

Step 2: Substitute a = 102 and b = 0 in the binomial theorem.

Step 3: Plug n = 5 into the equation.

Step 4: Calculate the value of (nCk)a^(n-k)b^k for k = 0, 1, 2, 3, 4, and 5.

Step 5: Add all the values of (nCk)a^(n-k)b^k together to get the result.

Answer: 10205025

Question:

Find (x+1)^6+(x−1)^6. Hence or otherwise evaluate (√2+1)^6+(√2−1)^6

Answer:

(x+1)^6+(x−1)^6

= (x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1) + (x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1)

= 2x^6 + 30x^4 + 30x^2 + 2

Now, substituting x = √2,

(√2+1)^6+(√2−1)^6

= 2(2)^6 + 30(2)^4 + 30(2)^2 + 2

= 128 + 480 + 120 + 2

= 630

Question:

Evaluate (96)^3

Answer:

Step 1: Multiply 96 by itself twice:

(96)^3 = 96 x 96 x 96

Step 2: Calculate 96 x 96:

(96)^3 = 96 x 96 x 96 = 9216 x 96

Step 3: Calculate 9216 x 96:

(96)^3 = 9216 x 96 = 8840192

Question:

Using Binomial Theorem, evaluate (101)^4

Answer:

Answer: Step 1: Using the Binomial Theorem, (a+b)^n = Σ(nCk)akbn-k, where n is the exponent and k is the index of summation.

Step 2: Substitute a = 101, b = 0, and n = 4 into the Binomial Theorem.

Step 3: Calculate the summation.

Step 4: (101)^4 = Σ(4Ck)101k0-k = Σ(4Ck)101k = 1 + 404 + 30,301 + 1,210,404.

Step 5: Therefore, (101)^4 = 1 + 404 + 30,301 + 1,210,404.

Question:

Expand the expression (2x−3)^6

Answer:

(2x−3)^6 = (2x)^6 − 6(2x)^5(3) + 15(2x)^4(3)^2 − 20(2x)^3(3)^3 + 15(2x)^2(3)^4 − 6(2x)(3)^5 + (3)^6

= 64x^6 - 1152x^53 + 5460x^43^2 - 8640x^33^3 + 5460x^23^4 - 1152x*3^5 + 3^6

Question:

Find (a+b)^4−(a−b)^4. Hence find s if (√3+√2)^4−(√3−√2)^4=40√s

Answer:

(a+b)^4−(a−b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4)

= 8a^3b + 24a^2b^2 + 16ab^3

Substituting a = √3 and b = √2,

(√3+√2)^4−(√3−√2)^4 = 8(√3)^3(√2) + 24(√3)^2(√2)^2 + 16(√3)(√2)^3

= 40√6

Hence, s = 6

Question:

Expand (x/3+1/x)^5

Answer:

Answer: Step 1: (x/3 + 1/x)^5 = Step 2: (x/3)^5 + 5(x/3)^4(1/x) + 10(x/3)^3(1/x)^2 + 10(x/3)^2(1/x)^3 + 5(x/3)(1/x)^4 + (1/x)^5 Step 3: x^5/243 + 5x^4/27 + 10x^3/9 + 10x^2/3 + 5x/x + 1/x^5 Step 4: x^5/243 + 5x^4/27 + 10x^3/9 + 10x^2/3 + 5 + 1/x^5

Question:

Expand (1−2x)^5

Answer:

(1−2x)^5 =

1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5

Question:

Expand (x+1/x)^6

Answer:

(x+1/x)^6 = (x^2 + 1)/x^2)^6 = (x^12 + 6x^10 + 15x^8 + 20x^6 + 15x^4 + 6x^2 + 1)/x^12 = x^0 + 6x^-2 + 15x^-4 + 20x^-6 + 15x^-8 + 6x^-10 + x^-12

Question:

Using Binomial theorem, evaluate (99)^5

Answer:

Answer: Step 1: Using Binomial theorem, (a+b)^n = (a^n) + (nC1)(a^(n-1)b) + (nC2)(a^(n-2)b^2) + …. + (nCn-1)(ab^(n-1)) + (b^n)

Step 2: Substitute a = 99 and b = 0 in the above equation.

Step 3: (99 + 0)^5 = (99^5) + (5C1)(99^40) + (5C2)(99^30^2) + (5C3)(99^20^3) + (5C4)(990^4) + (0^5)

Step 4: (99 + 0)^5 = (99^5) + (0) + (0) + (0) + (0) + (0)

Step 5: (99 + 0)^5 = (99^5)

Hence, the answer is 99^5 = 9,841,051,625.

Question:

Using binomial theorem, indicate which is larger (1.1)^10000 or 1000.

Answer:

1. Using the binomial theorem, we can expand (1.1)^10000 as:

(1.1)^10000 = (1 + 0.1)^10000 =

1 + 10000(0.1) + (10000)(10000-1)/2 (0.1)^2 + … + (10000)(0.1)^10000

2. Since the powers of 0.1 become increasingly small, we can ignore the terms after the first two and say that (1.1)^10000 is approximately equal to:

1 + 10000(0.1) = 1100

3. Since 1100 is greater than 1000, (1.1)^10000 is larger than 1000.

Question:

Find (a+b)^4−(a−b)^4. Hence, evaluate (√3+√2)^4−(√3−√2)^4.

Answer:

Step 1: (a+b)^4−(a−b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) − (a^4 − 4a^3b + 6a^2b^2 − 4ab^3 + b^4)

Step 2: (a+b)^4−(a−b)^4 = 8a^3b + 16a^2b^2 + 8ab^3

Step 3: Substituting a = √3 and b = √2, we get

(√3+√2)^4−(√3−√2)^4 = 8(√3)(√2)^3 + 16(√3)^2(√2)^2 + 8(√3)^3(√2)

Step 4: Simplifying further, we get

(√3+√2)^4−(√3−√2)^4 = 24√6 + 32√12 + 24√3

Step 5: Hence, the required answer is

(√3+√2)^4−(√3−√2)^4 = 80√3

Question:

Show that 9^(n+1)−8n−9 is divisible by 64, whenever n is a positive integer

Answer:

1. 9^(n+1) – 8n – 9
2. 9^(n+1) – 8n – 9 = 9(9^n) – 8n – 9
3. 9(9^n) – 8n – 9 = 9(9^n) – 8n – 9 + 8n
4. 9(9^n) – 8n – 9 + 8n = 9(9^n) – 9 + 8n
5. 9(9^n) – 9 + 8n = 9(9^n) – 9 + 8(8)
6. 9(9^n) – 9 + 8(8) = 9(9^n) – 9 + 64
7. 9(9^n) – 9 + 64 = 64(9^n - 1/8)
8. 64(9^n - 1/8) is divisible by 64

Therefore, 9^(n+1)−8n−9 is divisible by 64, whenever n is a positive integer.