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Prove, by induction, that the sum of the first n natural numbers, 1 + 2 + 3 + ⋯ + n, is \(\frac{n(n + 1)}{2}\) - Leaving Cert Mathematics - Question 3 - 2014
Question 3
Prove, by induction, that the sum of the first n natural numbers, 1 + 2 + 3 + ⋯ + n, is \(\frac{n(n + 1)}{2}\). Hence, or otherwise, prove that the sum of the firs... show full transcript
Worked Solution & Example Answer:Prove, by induction, that the sum of the first n natural numbers, 1 + 2 + 3 + ⋯ + n, is \(\frac{n(n + 1)}{2}\) - Leaving Cert Mathematics - Question 3 - 2014
Step 1
Prove, by induction, that the sum of the first n natural numbers, 1 + 2 + 3 + ⋯ + n, is \(\frac{n(n + 1)}{2}\).
Answer
To prove this statement by induction, we follow these steps:
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Base Case: For (n = 1):
LHS = (1)
RHS = (\frac{1(1 + 1)}{2} = 1)
Thus, the base case holds.
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Inductive Step: Assume the statement is true for (n = k), i.e., [1 + 2 + 3 + \ldots + k = \frac{k(k + 1)}{2}]
We must prove for (n = k + 1):
LHS = (1 + 2 + 3 + \ldots + k + (k + 1))
Using the inductive hypothesis, we find:
[ \text{LHS} = \frac{k(k + 1)}{2} + (k + 1) ]
Simplifying gives:
[ \text{LHS} = \frac{k(k + 1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2}]
This matches the RHS for (n = k + 1), hence the inductive step holds.
Therefore, by the principle of induction, the statement is true for all natural numbers (n).
Step 2
Hence, or otherwise, prove that the sum of the first n even natural numbers, 2 + 4 + 6 + ⋯ + 2n, is \(n^2 + n\).
Answer
To find the sum of the first (n) even natural numbers:
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Using the sum of the first n natural numbers:
The first n even numbers can be expressed as: [ 2 + 4 + 6 + \ldots + 2n = 2(1 + 2 + 3 + \ldots + n) ]
From part (a), we know: [ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} ]
Therefore: [ S_n = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) = n^2 + n]
Thus, the sum of the first n even natural numbers is confirmed as (n^2 + n).
Step 3
Using the results from (a) and (b) above, find an expression for the sum of the first n odd natural numbers in its simplest form.
Answer
To derive the sum of the first n odd natural numbers, we can use the results from (a) and (b):
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Identify how odd numbers relate to even numbers:
The first n odd natural numbers can be expressed as: [ 1 + 3 + 5 + \ldots + (2n - 1) ]
This can also be represented as: [ S_{odd} = S_{even} - n ] where (S_{even}) is the sum of the first n even numbers.
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Using the formula for the sum of even numbers:
From part (b), the sum of the first n even numbers is (n^2 + n), so:
[ S_{odd} = (n^2 + n) - n = n^2 ]
Therefore, we can conclude that the sum of the first n odd natural numbers is simply: [ S_{odd} = n^2 ].