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(a) (i) Prove by induction that, for any n, the sum of the first n natural numbers is \( \frac{n(n+1)}{2} \) - Leaving Cert Mathematics - Question 2 - 2012
Question 2
(a) (i) Prove by induction that, for any n, the sum of the first n natural numbers is \( \frac{n(n+1)}{2} \). (a) (ii) Find the sum of all the natural numbers from ... show full transcript
Worked Solution & Example Answer:(a) (i) Prove by induction that, for any n, the sum of the first n natural numbers is \( \frac{n(n+1)}{2} \) - Leaving Cert Mathematics - Question 2 - 2012
Step 1
Prove by induction that, for any n, the sum of the first n natural numbers is \( \frac{n(n+1)}{2} \)
Answer
To prove this statement by induction, we follow these steps:
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Base Case: When ( n = 1 ), the sum of the first natural number is 1. Calculating using the formula: [ \frac{1(1+1)}{2} = \frac{2}{2} = 1 ] Thus, the base case holds.
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Inductive Step: Assume the formula is true for some ( n = k ), i.e., [ S_k = \frac{k(k+1)}{2} ]
Now, we need to prove it for ( n = k + 1 ): [ S_{k+1} = S_k + (k + 1) = \frac{k(k+1)}{2} + (k + 1) ] Combine the terms: [ = \frac{k(k+1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2} ] This matches the formula for ( n = k + 1 ).
Thus, by mathematical induction, the statement is true for all natural numbers ( n ).
Step 2
Find the sum of all the natural numbers from 51 to 100, inclusive.
Answer
To find the sum of the natural numbers from 51 to 100, we can use the formula for the sum of the first( n ) natural numbers:
[ S_n = \frac{n(n+1)}{2} ]
First, calculate the total sum up to 100: [ S_{100} = \frac{100(101)}{2} = 5050 ]
Next, calculate the sum of the first 50 natural numbers: [ S_{50} = \frac{50(51)}{2} = 1275 ]
To find the sum from 51 to 100, subtract the sum of the first 50 from the sum up to 100: [ 51 + 52 + ... + 100 = S_{100} - S_{50} = 5050 - 1275 = 3775 ]
Therefore, the sum of all natural numbers from 51 to 100 is 3775.
Step 3
Given that p = log_e x, express log_e sqrt{x} + log_e (cx) in terms of p.
Answer
We start with: [ \log_e \sqrt{x} = \log_e x^{1/2} = \frac{1}{2} \log_e x = \frac{1}{2} p ]
Next, for ( \log_e (cx) ): [ \log_e (cx) = \log_e c + \log_e x ]
Using the fact that ( \log_e x = p ), we have: [ \log_e (cx) = \log_e c + p ]
Now, combining both results: [ \log_e \sqrt{x} + \log_e (cx) = \frac{1}{2} p + \log_e c + p = \frac{3}{2} p + \log_e c ]
Therefore, the expression in terms of p is:\n[ \log_e \sqrt{x} + \log_e (cx) = \frac{3}{2} p + \log_e c ]