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Problem 1: Prove that for all n ≥ 1, 1 + 3 + 5 + ... + (2n - 1) = n².
Proof by Mathematical Induction:
Base Case (n = 1):
Left-hand side (LHS): 1
Right-hand side (RHS): 1² = 1
Since LHS = RHS, the base case holds.
Inductive Step:
Assume the statement is true for some arbitrary positive integer k, i.e.,
1 + 3 + 5 + ... + (2k - 1) = k². *(Inductive Hypothesis)*
We need to prove it is true for k+1, i.e.,
1 + 3 + 5 + ... + (2k - 1) + (2(k+1) - 1) = (k+1)².
Starting with the left-hand side for k+1:
= [1 + 3 + 5 + ... + (2k - 1)] + (2(k+1) - 1)
= k² + (2k + 2 - 1) *(by the inductive hypothesis)*
= k² + (2k + 1)
= k² + 2k + 1
= (k + 1)².
This equals the right-hand side for k+1.
Therefore, if the statement is true for k, it is also true for k+1.
By the principle of mathematical induction, the identity 1 + 3 + 5 + ... + (2n - 1) = n² holds for all integers n ≥ 1.
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Problem 2: Prove that for every n ≥ 1, 1/(1·4) + 1/(4·7) + ... + 1/((3n-2)(3n+1)) = n/(3n+1).
Proof by Mathematical Induction:
Base Case (n = 1):
Left-hand side (LHS): 1/(1·4) = 1/4
Right-hand side (RHS): 1/(3·1 + 1) = 1/4
Since LHS = RHS, the base case holds.
Inductive Step:
Assume the statement is true for some arbitrary positive integer k, i.e.,
1/(1·4) + 1/(4·7) + ... + 1/((3k-2)(3k+1)) = k/(3k+1). *(Inductive Hypothesis)*
We need to prove it is true for k+1, i.e.,
1/(1·4) + 1/(4·7) + ... + 1/((3k-2)(3k+1)) + 1/((3(k+1)-2)(3(k+1)+1)) = (k+1)/(3(k+1)+1).
Simplify the new term:
(3(k+1)-2) = 3k + 3 - 2 = 3k + 1
(3(k+1)+1) = 3k + 3 + 1 = 3k + 4
So the new term is 1/((3k+1)(3k+4)).
Starting with the left-hand side for k+1:
= [1/(1·4) + 1/(4·7) + ... + 1/((3k-2)(3k+1))] + 1/((3k+1)(3k+4))
= k/(3k+1) + 1/((3k+1)(3k+4)) *(by the inductive hypothesis)*
Combine the terms over a common denominator (3k+1)(3k+4):
= [k(3k+4) + 1] / [(3k+1)(3k+4)]
= [3k² + 4k + 1] / [(3k+1)(3k+4)]
Factor the numerator:
3k² + 4k + 1 = (3k + 1)(k + 1)
So the expression becomes:
= [(3k + 1)(k + 1)] / [(3k + 1)(3k + 4)]
= (k + 1) / (3k + 4) *(canceling the common factor (3k+1), since 3k+1 ≠ 0 for k ≥ 1)*
Note that 3k + 4 = 3(k+1) + 1.
Thus, the result is (k+1)/(3(k+1)+1), which is the right-hand side for k+1.
Therefore, if the statement is true for k, it is also true for k+1.
By the principle of mathematical induction, the identity 1/(1·4) + 1/(4·7) + ... + 1/((3n-2)(3n+1)) = n/(3n+1) holds for all integers n ≥ 1.
Parent Tip: Review the logic above to help your child master the concept of mathematical induction worksheet.
