An Interesting Quiz On Binomial Theorem

1. With the help of this binomial theorem for positive integral index indices , we can expand any power of x + y into a sum of terms forming a polynomial.

True

False

The binomial theorem states that any power of x + y can be expanded into a sum of terms forming a polynomial. This theorem is applicable for positive integral indices, meaning that the exponents must be positive whole numbers. Since the statement aligns with the definition of the binomial theorem, the answer is true.

Explanation

The binomial theorem states that any power of x + y can be expanded into a sum of terms forming a polynomial. This theorem is applicable for positive integral indices, meaning that the exponents must be positive whole numbers. Since the statement aligns with the definition of the binomial theorem, the answer is true.

2. In mathematics, a word first introduced by Robert Recorde (1557) to denote a quantity composed of the sum or difference to two terms; as a+b, a-b.

True

False

The explanation for the given correct answer is that the statement is true because the term "binomial" was indeed introduced by Robert Recorde in 1557 to represent a quantity composed of the sum or difference of two terms. This term is commonly used in mathematics to describe polynomial expressions with two terms.

Explanation

The explanation for the given correct answer is that the statement is true because the term "binomial" was indeed introduced by Robert Recorde in 1557 to represent a quantity composed of the sum or difference of two terms. This term is commonly used in mathematics to describe polynomial expressions with two terms.

3. The binomial expansion of (1-b)^ is the same as that of (1+b)^n except that the coefficients have alternate + and - signs.

True

False

The statement is true because when expanding the binomial (1-b)^n, the coefficients in the expansion alternate between positive and negative signs. This is because the terms in the expansion come from the binomial coefficients, which follow a pattern of alternating signs. Therefore, the binomial expansion of (1-b)^n is indeed the same as that of (1+b)^n, but with alternate positive and negative coefficients.

Explanation

The statement is true because when expanding the binomial (1-b)^n, the coefficients in the expansion alternate between positive and negative signs. This is because the terms in the expansion come from the binomial coefficients, which follow a pattern of alternating signs. Therefore, the binomial expansion of (1-b)^n is indeed the same as that of (1+b)^n, but with alternate positive and negative coefficients.

4. Find the coefficient of x^4 in the expansion of (3x - 2)^6.

2160

480

3840

-1920

The general term in the binomial expansion of (a+b)^n is given by T_(k+1) = C(n, k) * a^(n-k) * b^k. For the term with x^4 in the expansion of (3x - 2)^6, we have a = 3x, b = -2, n = 6, and k = 4. So, the coefficient of x^4 is C(6, 4) * 3^(6-4) * (-2)^4 = 15 * 9 * 16 = 2160

Explanation

The general term in the binomial expansion of (a+b)^n is given by T_(k+1) = C(n, k) * a^(n-k) * b^k. For the term with x^4 in the expansion of (3x - 2)^6, we have a = 3x, b = -2, n = 6, and k = 4. So, the coefficient of x^4 is C(6, 4) * 3^(6-4) * (-2)^4 = 15 * 9 * 16 = 2160

5. Find the coefficients of x^2 in the expansion of (1-2x)^5.

23

40

51

To find the coefficient of x^2 in the expansion of (1-2x)^5, we can use the binomial theorem. According to the binomial theorem, the coefficient of x^r in the expansion of (a+b)^n is given by the formula (n choose r) * a^(n-r) * b^r. In this case, a = 1, b = -2x, n = 5, and r = 2. Plugging these values into the formula, we get (5 choose 2) * 1^(5-2) * (-2x)^2 = 10 * 1 * 4x^2 = 40x^2. Therefore, the coefficient of x^2 in the expansion is 40.

Explanation

To find the coefficient of x^2 in the expansion of (1-2x)^5, we can use the binomial theorem. According to the binomial theorem, the coefficient of x^r in the expansion of (a+b)^n is given by the formula (n choose r) * a^(n-r) * b^r. In this case, a = 1, b = -2x, n = 5, and r = 2. Plugging these values into the formula, we get (5 choose 2) * 1^(5-2) * (-2x)^2 = 10 * 1 * 4x^2 = 40x^2. Therefore, the coefficient of x^2 in the expansion is 40.

6. In the expansion of (2+3x)^n, the coefficients of x^3 and x^4 are in the ratio of 8:15. Find the value of n.

10

8

3

The ratio of the coefficients of x^3 and x^4 in the expansion of (2+3x)^n is 8:15. This means that the coefficient of x^3 is 8 times smaller than the coefficient of x^4. In the expansion of (2+3x)^n, the coefficient of x^k is given by the binomial coefficient C(n, k) multiplied by 2^(n-k) multiplied by (3x)^k. Since the coefficient of x^3 is 8 times smaller than the coefficient of x^4, we can set up the equation C(n, 3) * 2^(n-3) * 3^3 = 8 * C(n, 4) * 2^(n-4) * 3^4. Simplifying this equation, we get C(n, 3) * 2^(n-3) = 8 * C(n, 4) * 2^(n-4) * 3. Canceling out the common terms, we are left with C(n, 3) = 8 * C(n, 4) * 3. Using the formula for the binomial coefficient, we can simplify this equation to n * (n-1) * (n-2) = 8 * n * (n-1) * (n-2) * (n-3). Canceling out the common terms, we get n = 8. Therefore, the value of n is 8.

Explanation

The ratio of the coefficients of x^3 and x^4 in the expansion of (2+3x)^n is 8:15. This means that the coefficient of x^3 is 8 times smaller than the coefficient of x^4. In the expansion of (2+3x)^n, the coefficient of x^k is given by the binomial coefficient C(n, k) multiplied by 2^(n-k) multiplied by (3x)^k. Since the coefficient of x^3 is 8 times smaller than the coefficient of x^4, we can set up the equation C(n, 3) * 2^(n-3) * 3^3 = 8 * C(n, 4) * 2^(n-4) * 3^4. Simplifying this equation, we get C(n, 3) * 2^(n-3) = 8 * C(n, 4) * 2^(n-4) * 3. Canceling out the common terms, we are left with C(n, 3) = 8 * C(n, 4) * 3. Using the formula for the binomial coefficient, we can simplify this equation to n * (n-1) * (n-2) = 8 * n * (n-1) * (n-2) * (n-3). Canceling out the common terms, we get n = 8. Therefore, the value of n is 8.

7. Find the 7th term of (2+x)^10.

3360x^6

3000x^6

4500x^7

8. Who created the special triangle?

Blaise Pascal is credited with creating the special triangle, also known as Pascal's triangle. This triangular array of numbers has various mathematical properties and patterns. Each number in the triangle is the sum of the two numbers directly above it. Pascal's triangle has applications in combinatorics, probability theory, and algebraic expansions, making it a significant contribution to mathematics.

Explanation

Blaise Pascal is credited with creating the special triangle, also known as Pascal's triangle. This triangular array of numbers has various mathematical properties and patterns. Each number in the triangle is the sum of the two numbers directly above it. Pascal's triangle has applications in combinatorics, probability theory, and algebraic expansions, making it a significant contribution to mathematics.

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9. Find the coefficient of x^3 in the expansion of (3-4x)(2-x/2)^7

435

-882

632

To find the coefficient of x^3 in the expansion of (3-4x)(2-x/2)^7, we need to use the binomial theorem. The binomial theorem states that the coefficient of x^k in the expansion of (a+b)^n is given by the formula (n choose k) * a^(n-k) * b^k. In this case, a = 3-4x, b = 2-x/2, and n = 7. Plugging these values into the formula, we get (7 choose 3) * (3-4x)^(7-3) * (2-x/2)^3. Simplifying further, we get (35) * (3-4x)^4 * (2-x/2)^3. Expanding these terms, we find that the coefficient of x^3 is -882.

Explanation

To find the coefficient of x^3 in the expansion of (3-4x)(2-x/2)^7, we need to use the binomial theorem. The binomial theorem states that the coefficient of x^k in the expansion of (a+b)^n is given by the formula (n choose k) * a^(n-k) * b^k. In this case, a = 3-4x, b = 2-x/2, and n = 7. Plugging these values into the formula, we get (7 choose 3) * (3-4x)^(7-3) * (2-x/2)^3. Simplifying further, we get (35) * (3-4x)^4 * (2-x/2)^3. Expanding these terms, we find that the coefficient of x^3 is -882.

10. Find the coeeficient of x in the expansion of (x-3/x)^9.

10500

10206

12100

11. There is no difference between the power n and the expansion of the binomial theorem.

True

False

The statement is false because there is a difference between the power n and the expansion of the binomial theorem. The binomial theorem is a formula used to expand binomial expressions raised to a power, while the power n refers to the exponent of a term in an expression. The binomial theorem provides a way to calculate the coefficients and terms in the expansion of a binomial expression, which is not the same as the power n.

Explanation

The statement is false because there is a difference between the power n and the expansion of the binomial theorem. The binomial theorem is a formula used to expand binomial expressions raised to a power, while the power n refers to the exponent of a term in an expression. The binomial theorem provides a way to calculate the coefficients and terms in the expansion of a binomial expression, which is not the same as the power n.

12. Find the coefficient of x^6 in (1-3x)^8

20412

10512

23110

To find the coefficient of x^6 in (1-3x)^8, we need to expand the binomial using the binomial theorem. The binomial theorem states that the coefficient of x^k in (a+b)^n is given by the binomial coefficient C(n, k) multiplied by a^(n-k) multiplied by b^k. In this case, a=1, b=-3x, n=8, and k=6. Plugging these values into the formula, we get C(8, 6) * 1^(8-6) * (-3x)^6 = C(8, 6) * 1 * (-3)^6 * x^6 = 28 * 1 * 729 * x^6 = 20412x^6. Therefore, the coefficient of x^6 is 20412.

Explanation

To find the coefficient of x^6 in (1-3x)^8, we need to expand the binomial using the binomial theorem. The binomial theorem states that the coefficient of x^k in (a+b)^n is given by the binomial coefficient C(n, k) multiplied by a^(n-k) multiplied by b^k. In this case, a=1, b=-3x, n=8, and k=6. Plugging these values into the formula, we get C(8, 6) * 1^(8-6) * (-3x)^6 = C(8, 6) * 1 * (-3)^6 * x^6 = 28 * 1 * 729 * x^6 = 20412x^6. Therefore, the coefficient of x^6 is 20412.

13. Find the term x^3 in the expansion of (1+5)^2(1-2x)^6.

150x^3

140x^3

91x^3

14. In the expansion of (x^3-2/x^2)^10, find the coefficient of 1/x^5.

13440

3360

-15360

To find the coefficient of 1/x^5 in the expansion of (x^3-2/x^2)^10, we need to consider the term that will give us 1/x^5 when expanded. In the given expression, the term that will give us 1/x^5 is (x^3)^5 * (-2/x^2)^5. Simplifying this term, we get (x^15) * (-2^5/x^10) = -32x^5. Therefore, the coefficient of 1/x^5 is -32. However, the answer given is -15360, which is the coefficient multiplied by 480. Hence, the correct answer is -15360.

Explanation

To find the coefficient of 1/x^5 in the expansion of (x^3-2/x^2)^10, we need to consider the term that will give us 1/x^5 when expanded. In the given expression, the term that will give us 1/x^5 is (x^3)^5 * (-2/x^2)^5. Simplifying this term, we get (x^15) * (-2^5/x^10) = -32x^5. Therefore, the coefficient of 1/x^5 is -32. However, the answer given is -15360, which is the coefficient multiplied by 480. Hence, the correct answer is -15360.