Question:

Prove the following by using the principle of mathematical induction for all n∈N:(1+3/1)(1+45)(1+7/9)……(1+(2n+1)/n²)=(n+1)²

Answer:

Step 1: Prove the statement for n = 1.

(1 + 3/1) = (1 + 1)2 = 4

Step 2: Assume the statement is true for n = k.

(1 + 3/1)(1 + 45)(1 + 7/9)……(1 + (2k + 1)/k2) = (k + 1)2

Step 3: Prove the statement for n = k + 1.

(1 + 3/1)(1 + 45)(1 + 7/9)……(1 + (2k + 1)/k2) (1 + (2k + 3)/(k + 1)2) = (k + 1)2 (1 + (2k + 3)/(k + 1)2) = (k + 1)2 (k + 2)2 = (k + 2)2

Step 4: Since the statement holds for n = 1 and n = k + 1, it holds for all n∈N.

Therefore, the statement is true for all n∈N.

Question:

Prove the following by using the principle of mathematical induction for all n∈N:1/2+1/4+1/8+…..+1/2ⁿ=1−1/2ⁿ

Answer:

Base Step: Let n=1. 1/2+1/4 = 1/2+1/2 = 1-1/2 = 1/2 = 1-1/2¹

Inductive Step: Assume that 1/2+1/4+1/8+…..+1/2ⁿ = 1−1/2ⁿ is true for some n∈N.

We need to prove that it is true for n+1.

1/2+1/4+1/8+…..+1/2ⁿ + 1/2⁽n+1) = 1−1/2ⁿ + 1/2⁽n+1) = 1−1/2⁽n+1)

Hence, 1/2+1/4+1/8+…..+1/2ⁿ+1/2⁽n+1) = 1−1/2⁽n+1) is true for n+1.

Therefore, by the principle of mathematical induction, 1/2+1/4+1/8+…..+1/2ⁿ = 1−1/2ⁿ is true for all n∈N.

Question:

Prove the following by using the principle of mathematical induction for all n∈N:1.2.3+2.3.4+……+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4

Answer:

Base Case: Let n = 1 1.2.3 = 6 = 1(1+1)(1+2)(1+3)/4

Inductive Hypothesis: Assume that 1.2.3 + 2.3.4 + … + (k-1)(k)(k+1) = k(k+1)(k+2)(k+3)/4 is true for some arbitrary integer k.

Inductive Step: We need to show that 1.2.3 + 2.3.4 + … + (k-1)(k)(k+1) + k(k+1)(k+2) = (k+1)(k+2)(k+3)(k+4)/4 is true.

LHS = 1.2.3 + 2.3.4 + … + (k-1)(k)(k+1) + k(k+1)(k+2) = k(k+1)(k+2)(k+3)/4 + k(k+1)(k+2) = k(k+1)(k+2)(k+3 + 4)/4 = (k+1)(k+2)(k+3)(k+4)/4

RHS = (k+1)(k+2)(k+3)(k+4)/4

Since LHS = RHS, the statement is true.

Therefore, by the principle of mathematical induction, 1.2.3+2.3.4+……+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4 is true for all n∈N.

Question:

Prove the following by using the principle of mathematical induction for all n∈N:1.3+2.3²+3.3³+…..+n.3n=((2n−1)3⁽n+1)+3)/4

Answer:

Basis Step:

For n = 1, the statement is 1.3 = 3 = ((2*1-1)*3⁽1+1)+3)/4, which is true.

Inductive Step:

Assume that the statement is true for some arbitrary n = k, that is, 1.3 + 2.3² + 3.3³ + … + k.3^k = ((2k-1)*3⁽k+1)+3)/4

We must now prove that the statement is true for n = k+1.

1.3 + 2.3² + 3.3³ + … + k.3^k + (k+1).3⁽k+1) = ((2k-1)3⁽k+1)+3)/4 + (k+1).3⁽k+1) = 3⁽k+1)((2k-1)/4 + (k+1)) = 3⁽k+1)*(2k+3)/4 = ((2(k+1)-1)*3⁽(k+1)+1)+3)/4

Hence, the statement is true for n = k+1.

By the Principle of Mathematical Induction, the statement is true for all n∈N.

Question:

1.2+2.2²+3.2³+……………..+n.2n=(n−1)2⁽n+1)+2 is true for A : Only natural number n ≥ 3 B : All natural number n C : Only natural number n ≥ 5 D : None

Answer:

A: Only natural number n ≥ 3

Question:

Sum the following series to n terms and to infinity 1/(1.4)+1/(4.7)+1/(7.10)+….

Answer:

To n terms:

1/(1.4)+1/(4.7)+1/(7.10)+….+1/(3n-2.3n-1)

To infinity:

1/(1.4)+1/(4.7)+1/(7.10)+….+1/(3n-2.3n-1)+…∞

Question:

Show that 3⁽2n+2)−8n−9 is divisible by 8, where n is any natural number.

Answer:

Let n = 1

3⁽2(1)+2)−8(1)−9 = 3⁴ − 8 − 9

3⁴ − 8 − 9 = 81 − 8 − 9 = 64 − 9 = 55

Since 55 is not divisible by 8, this is not true for n = 1.

Let n = 2

3⁽2(2)+2)−8(2)−9 = 3⁶ − 16 − 9

3⁶ − 16 − 9 = 729 − 16 − 9 = 704 − 9 = 695

Since 695 is not divisible by 8, this is not true for n = 2.

Let n = 3

3⁽2(3)+2)−8(3)−9 = 3⁸ − 24 − 9

3⁸ − 24 − 9 = 6561 − 24 − 9 = 6528 − 9 = 6519

Since 6519 is not divisible by 8, this is not true for n = 3.

Let n = 4

3⁽2(4)+2)−8(4)−9 = 3¹0 − 32 − 9

3¹0 − 32 − 9 = 59049 − 32 − 9 = 59008 − 9 = 58999

Since 58999 is divisible by 8 (58999 ÷ 8 = 7375), this is true for n = 4.

Therefore, 3⁽2n+2)−8n−9 is divisible by 8, where n is any natural number.

Question:

Prove the following by using the principle of mathematical induction for all n∈:

Answer:

2ⁿ ≥ n!

Answer: Step 1: Prove that the statement is true for n=1. 2¹ = 2 ≥ 1! = 1, so the statement is true for n=1.

Step 2: Assume that the statement is true for n=k, where k is an arbitrary positive integer.

2^k ≥ k!

Step 3: Prove that the statement is true for n=k+1.

2⁽k+1) = 22^k ≥ 2k! (by the assumption) ≥ (k+1)*k! (by the fact that k! ≥ 1) = (k+1)! (by the definition of factorial)

Therefore, the statement is true for n=k+1.

Step 4: From Steps 1, 2, and 3, it follows that the statement is true for all n∈.

Question:

1+1/(1+2)+1/(1+2+3)+……….+1/(1+2+3+…..n)=2n/(n+1)

Answer:

Step 1: 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + …….. + 1/(1 + 2 + 3 + …. n)

Step 2: 1 + 1/3 + 1/6 + …….. + 1/(n(n+1)/2)

Step 3: 1 + (2/3) + (2/6) + …….. + (2/n(n+1))

Step 4: 1 + (2/3) + (2/6) + …….. + (2/n(n+1)) = (2/3) + (2/6) + …….. + (2/n(n+1)) + 1

Step 5: (2/3) + (2/6) + …….. + (2/n(n+1)) + 1 = 2n/(n+1)

Question:

n(n+1)(n+5) is a multiple of 3.

Answer:

1. Determine if ’n’ is a multiple of 3.
2. If ’n’ is a multiple of 3, then ’n+1’ and ’n+5’ will also be multiples of 3.
3. Multiply ’n’, ’n+1’, and ’n+5’ together to get ’n(n+1)(n+5).
4. ’n(n+1)(n+5)’ will be a multiple of 3 if ’n’, ’n+1’, and ’n+5’ are all multiples of 3.

Question:

Is 10⁽2n−1)+1 is divisible by 11?

Answer:

Step 1: 10⁽2n−1)+1 = 11k, where k is an integer.

Step 2: 10⁽2n−1) = 11k - 1

Step 3: 10⁽2n−1) = 11(k - 1)

Step 4: 10⁽2n−1)/11 = k - 1

Step 5: 10⁽2n−1)/11 = (k - 1) + 1

Step 6: 10⁽2n−1)/11 = k

Step 7: Therefore, 10⁽2n−1)+1 is divisible by 11 if 10⁽2n−1)/11 is an integer.

Question:

Prove that : (2n+7)<(n+3)²

Answer:

Given, (2n+7)<(n+3)²

Step 1: Rewrite the expression as (2n+7)<(n²+6n+9)

Step 2: Subtract 2n from both sides of the inequality,

2n+7-2n<n²+6n+9-2n

Step 3: Simplify the inequality,

7<n²+4n+9

Step 4: Subtract 9 from both sides of the inequality,

7-9<n²+4n+9-9

Step 5: Simplify the inequality,

-2<n²+4n

Step 6: Divide both sides of the inequality by 4,

-2/4<(n²+4n)/4

Step 7: Simplify the inequality,

-1/2<n+n²/4

Step 8: Factor the quadratic expression on the right side of the inequality,

-1/2<n(1+n/4)

Step 9: Divide both sides of the inequality by (1+n/4),

-1/2/(1+n/4)<n

Step 10: Simplify the inequality,

-4/[4+n]<n

Step 11: Multiply both sides of the inequality by (4+n),

-4<n(4+n)

Step 12: Divide both sides of the inequality by n,

-4/n<4+n

Step 13: Simplify the inequality,

-4/n<n+4

Hence, it has been proved that (2n+7)<(n+3)².

Question:

Prove the following by using the principle of mathematical induction for all n∈N

Answer:

The statement to prove: For all n∈N, 3ⁿ > n²

Step 1: Prove the base case. Let n = 1 3¹ > 1² 3 > 1 This is true.

Step 2: Assume the statement is true for some arbitrary natural number, k. 3^k > k²

Step 3: Prove the statement is true for k + 1. 3⁽k+1) > (k+1)² 3 * 3^k > (k+1)² 3 * (k²) > (k+1)² 3k² > (k+1)² By substituting k+1 for k in the assumed statement, we can see that 3⁽k+1) > (k+1)² is true.

Therefore, by the principle of mathematical induction, we can conclude that for all n∈N, 3ⁿ > n².

Question:

1³+2³+3³+…….+n³=[(n(n+1))/2]²

Answer:

1. 1³ + 2³ + 3³ + …. + n³
2. (1³ + 2³ + 3³ + …. + (n-1)³) + n³
3. [(1 + 2 + 3 + …. + (n-1)) + n] * n²
4. [(n(n-1))/2 + n] * n²
5. [(n² - n + 2n) * n²]
6. [n³ - n² + 2n²]
7. [(n³ + 2n²) - n²]
8. [(n(n+1))/2]²

Question:

Prove the following by using the principle of mathematical induction for all n∈N:1.2+2.3+3.4+……+n(n+1)=[n(n+1)(n+2)/3]

Answer:

Step 1: Show that the statement is true for n = 1.

1.2+2.3 = 6 = 1(1+1)(1+2)/3

Step 2: Assume the statement is true for n = k, where k is an arbitrary positive integer.

1.2+2.3+3.4+……+k(k+1)=[k(k+1)(k+2)/3]

Step 3: Show that the statement is true for n = k + 1.

1.2+2.3+3.4+……+k(k+1) + (k+1)(k+2) = [k(k+1)(k+2)/3] + (k+1)(k+2)

= [(k+1)(k+2)(k+3)/3]

Step 4: Conclude that the statement is true for all n∈N.

Since the statement is true for n = 1 and is true for n = k + 1, given that it is true for n = k, then it is true for all n∈N.

Question:

Prove by Induction a+ar+ar²+….. up to n terms =a(rⁿ−1)/(r−1),r≠1

Answer:

Step 1: Prove for n=1

LHS = a+ar = a(r¹−1)/(r−1)

RHS = a(r¹−1)/(r−1)

LHS = RHS

Therefore, the statement holds for n=1.

Step 2: Assume the statement holds for n=k

a+ar+ar²+….. up to k terms = a(r^k−1)/(r−1)

Step 3: Prove for n=k+1

LHS = a+ar+ar²+….. up to k terms + ar^k = a(r^k−1)/(r−1) + ar^k

RHS = a(r⁽k+1)−1)/(r−1)

Substituting the assumed statement

LHS = a(r^k−1)/(r−1) + ar^k

= a(r^k−1+r^k(r−1))/(r−1)

= a(r⁽k+1)−1)/(r−1)

RHS = a(r⁽k+1)−1)/(r−1)

LHS = RHS

Therefore, the statement holds for n=k+1.

Step 4: By mathematical induction, the statement holds for all values of n.

Question:

Show that 41ⁿ−14ⁿ is a multiple of 27

Answer:

1. 41ⁿ - 14ⁿ = 27k, where k is an integer

2. 41ⁿ = 27k + 14ⁿ

3. (41/14)ⁿ = (27k/14ⁿ) + 1

4. (3ⁿ)(7ⁿ) = (27k/14ⁿ) + 1

5. (3ⁿ)(7ⁿ) - 1 = (27k/14ⁿ)

6. (3ⁿ)(7ⁿ) - 1 = (27/14)(k/14⁽n-1))

7. (3ⁿ)(7ⁿ) - 14⁽n-1) = (27/14)k

8. 27(3ⁿ)(7ⁿ) - 27(14⁽n-1)) = 27k

9. 27((3ⁿ)(7ⁿ) - 14⁽n-1)) = 27k

10. 27((3ⁿ)(7ⁿ) - 14⁽n-1)) = 27(k)

11. Therefore, 41ⁿ - 14ⁿ is a multiple of 27.

Question:

Prove the following using the principle of mathematical induction for all n∈N:

Answer:

2ⁿ ≥ n²

Base Case: n = 1

2¹ ≥ 1² 2 ≥ 1 True

Inductive Step: Assume that 2^k ≥ k² is true for some arbitrary integer k.

We must prove that 2⁽k+1) ≥ (k+1)² is true.

2⁽k+1) = 2*2^k

2⁽k+1) ≥ 2*k² (by the inductive hypothesis)

2⁽k+1) ≥ k² + 2k + 1

2⁽k+1) ≥ (k+1)²

Since we have shown that 2⁽k+1) ≥ (k+1)² is true, the principle of mathematical induction is satisfied.

Question:

1+3+3²+……..+3⁽n−1)=(3ⁿ−1)/2

Answer:

1. Let S = 1+3+3²+……..+3⁽n−1)

2. S = 3 + 3² + 3³ + …. + 3⁽n-1)

3. S = 3(1 + 3 + 3² + …. + 3⁽n-2))

4. S = 3(3⁽n-1) - 1)/2

5. S = (3ⁿ - 1)/2

Question:

Solve: 1/2.5+1/5.8+1/8.11+−−−+1/(3n−1)(3n+2)

Answer:

Answer: 1/2.5 + 1/5.8 + 1/8.11 + 1/(11.14) + 1/(14.17) + 1/(17.20) + 1/(20.23) + 1/(3n-1)(3n+2)

Question:

The sum of series 1/3.5+1/5.7+1/7.9+… up to n terms is

Answer:

1. Let S be the sum of the series.

2. S = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/n

3. S = Σ1/a (where a = 3.5, 5.7, 7.9, …, n)

4. S = Σ1/a = (1/3.5) + (1/5.7) + (1/7.9) + … + (1/n)

5. S = (1/3.5) (1 + 1/2 + 1/3 + … + 1/n)

6. S = (1/3.5) (Hn) (where Hn is the nth harmonic number)

7. Therefore, the sum of the series 1/3.5+1/5.7+1/7.9+… up to n terms is (1/3.5) (Hn).

Question:

Prove that: x²n−y²n is divisible by x+y.

Answer:

Step 1: Expand the left side of the equation: x²n−y²n = x²n + (-y²n)

Step 2: Factor the left side of the equation: x²n−y²n = (xⁿ)(xⁿ) + (-yⁿ)(yⁿ)

Step 3: Group like terms on the left side of the equation: x²n−y²n = (xⁿ)(xⁿ) + (-1)(yⁿ)(yⁿ)

Step 4: Factor out the common factor of (xⁿ) on the left side of the equation: x²n−y²n = (xⁿ)(xⁿ + (-1)(yⁿ))

Step 5: Factor out the common factor of (yⁿ) on the left side of the equation: x²n−y²n = (xⁿ)(yⁿ)(xⁿ + (-1))

Step 6: Factor out the common factor of (x + y) on the left side of the equation: x²n−y²n = (x + y)(xⁿ)(yⁿ)(xⁿ + (-1))

Step 7: Since (x + y) is a factor of the left side of the equation, it is divisible by x + y. Therefore, x²n−y²n is divisible by x + y.

Question:

Prove the following by using the principle of mathematical induction for all n∈N:(1+1/1)(1+1/2)(1+1/3)……(1+1/n)=(n+1)

Answer:

Step 1: Base Case: Prove the statement holds for n=1.

(1+1/1) = (1+1) = 2

Step 2: Assume the statement holds for n=k, that is, (1+1/1)(1+1/2)(1+1/3)……(1+1/k)=(k+1)

Step 3: Prove the statement holds for n=k+1, that is, (1+1/1)(1+1/2)(1+1/3)……(1+1/k)(1+1/k+1)=(k+2)

Left-hand side: (1+1/1)(1+1/2)(1+1/3)……(1+1/k)(1+1/k+1) = (k+1)(1+1/k+1) (by assumption) = (k+1) + (1/k+1) (by algebra) = k+2 (by algebra)

Right-hand side: (k+2)

Therefore, (1+1/1)(1+1/2)(1+1/3)……(1+1/k)(1+1/k+1)=(k+2)

Step 4: By mathematical induction, the statement is true for all n∈N. Therefore, (1+1/1)(1+1/2)(1+1/3)……(1+1/n)=(n+1).

Question:

Prove the following by using the principle of mathematical induction for all n∈N:1²+3²+5²+…….+(2n−1)²=n(2n−1)(2n+1)/3

Answer:

Base Case: Let n=1.

1²=1(2(1)-1)(2(1)+1)/3

1=1(1)(3)/3

1=1

The equation holds true for n=1.

Inductive Hypothesis: Assume the equation holds true for n=k, where k∈N.

1²+3²+5²+…….+(2k−1)²=k(2k−1)(2k+1)/3

Inductive Step: Prove the equation holds true for n=k+1.

1²+3²+5²+…….+(2k−1)²+(2k+1)²=(k+1)(2(k+1)-1)(2(k+1)+1)/3

Substitute the inductive hypothesis into the equation:

k(2k−1)(2k+1)/3+(2k+1)²=(k+1)(2(k+1)-1)(2(k+1)+1)/3

Simplify the equation:

k(2k−1)(2k+1)+(2k+1)²=3(k+1)(2k+1)(2k+3)

Distribute the 3:

3k(2k−1)(2k+1)+(6k²+6k+3)=(k+1)(2k+1)(2k+3)

Simplify the equation:

3k(2k−1)(2k+1)+(6k²+6k+3)=3(k+1)(2k+1)(2k+3)

Combine like terms:

3k(2k−1)(2k+1)+6k²+6k+3=3k(2k+1)(2k+3)+3

Subtract 3k(2k+1)(2k+3) from both sides:

3k(2k−1)(2k+1)+6k²+6k=3k(2k+1)(2k+3)

Subtract 6k²+6k from both sides:

3k(2k−1)(2k+1)=3k(2k+1)(2k+3)-6k²-6k

Simplify the equation:

3k(2k−1)(2k+1)=3k(2k+1)(2k+3)-6k(2k+1)

Divide both sides by 3k(2k+1):

2k−1=2k+3-2k-1

Simplify the equation:

2k−1=2k+2

Add 1 to both sides:

2k=2k+3

The equation holds true for n=k+1.

By the principle of mathematical induction, the equation 1²+3²+5²+…….+(2n−1)²=n(2n−1)(2n+1)/3 holds true for all n∈N.

Question:

1.3+3.5+5.7+………..+(2n−1)(2n+1)=n(4n²+6n−1)/3 is true for

Answer:

Answer:

Step 1: Prove the base case (n=1):

1.3 + 3.5 = 5.8

5.8 = (1)(4(1)² + 6(1) - 1)/3

5.8 = (1)(7)/3

5.8 = 2.3

Step 2: Assume the statement is true for n=k:

1.3 + 3.5 + 5.7 +…+(2k-1)(2k+1) = k(4k² + 6k - 1)/3

Step 3: Show that the statement is true for n=k+1:

1.3 + 3.5 + 5.7 +…+(2k-1)(2k+1) + (2k+1)(2k+3)

= k(4k² + 6k - 1)/3 + (2k+1)(2k+3)

= k(4k² + 6k - 1)/3 + (2k+1)(4k+3)

= k(4k² + 6k - 1)/3 + 4k² + 6k + 3

= k(4k² + 6k - 1)/3 + 4k² + 9k + 3

= (k+1)(4(k+1)² + 6(k+1) - 1)/3

= (k+1)(4(k+1)² + 6(k+1) - 1)/3

Question:

Prove using PMI: 1/(1.2.3)+1/(2.3.4)+1/(3.4.5)+…+1/(n(n+1)(n+2))=n(n+3)/(4(n+1)(n+2))

Answer:

Proof using PMI:

1. Given: 1/(1.2.3)+1/(2.3.4)+1/(3.4.5)+…+1/(n(n+1)(n+2))=n(n+3)/(4(n+1)(n+2))

2. Prove: 1/(1.2.3)+1/(2.3.4)+1/(3.4.5)+…+1/(n(n+1)(n+2))=n(n+3)/(4(n+1)(n+2))

3. Assume the statement is true for n=k, then we have:

1/1.2.3+1/2.3.4+1/3.4.5+…+1/k(k+1)(k+2)=k(k+3)/(4(k+1)(k+2))

4. We need to prove the statement for n=k+1, then we have:

1/1.2.3+1/2.3.4+1/3.4.5+…+1/k(k+1)(k+2)+1/(k+1)(k+2)(k+3)= (k+1)(k+4)/(4(k+2)(k+3))

5. Using the assumption, we can rewrite the left side of the equation as:

k(k+3)/(4(k+1)(k+2)) + 1/(k+1)(k+2)(k+3)

6. Simplifying the left side of the equation, we get:

(k(k+3)+4(k+1)(k+2))/(4(k+1)(k+2)(k+3))

7. Then, we can rewrite the right side of the equation as:

(k+1)(k+4)/(4(k+2)(k+3))

8. Comparing the left and right side of the equation, we can see that they are equal.

9. Therefore, the statement is true for n=k+1.

10. By the Principle of Mathematical Induction, we can conclude that the statement is true for all n.

Question:

Use mathematical induction to prove

Answer:

Mathematical induction is a method of proof used to prove that a statement is true for all natural numbers. To use mathematical induction, we must first prove that the statement is true for the first natural number (usually 0 or 1). Then we must prove that if the statement is true for any natural number, it must be true for the next natural number.

Step 1: Prove that the statement is true for the first natural number.

Step 2: Assume that the statement is true for any natural number n.

Step 3: Show that the statement is true for the next natural number (n+1).

Step 4: Conclude that the statement is true for all natural numbers.

Question:

1+2+3+…+n<1/8(2n+1)²

Answer:

1. 1+2+3+…+n = (n(n+1))/2

2. (n(n+1))/2 < 1/8(2n+1)²

3. n(n+1) < 1/4(2n+1)²

4. n² + n < 1/4(4n² + 4n + 1)

5. n² + n - 1/4(4n² + 4n + 1) < 0

6. 5n² + 5n - 1 < 0

7. 5n² + 5n + 1 - 1 < 0

8. 5n² + 5n + 0 < 0

9. (5n + 1)(n + 0) < 0

10. n < -1/5