Photo AI
(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 the statement by induction, we proceed as follows:
Base Case: For ( n = 1 ), the sum of the first natural number is: [ 1 = \frac{1(1+1)}{2} ] This holds true.
Inductive Step: Assume the statement holds for some ( n \geq 1 ): [ 1 + 2 + \ldots + n = \frac{n(n+1)}{2} ] Now for ( n + 1 ): [ 1 + 2 + \ldots + n + (n + 1) = \frac{n(n + 1)}{2} + (n + 1) ] This simplifies to: [ = \frac{n(n + 1) + 2(n + 1)}{2} = \frac{(n + 1)(n + 2)}{2} ] Thus, the statement is true for ( n + 1 ). Therefore, by mathematical induction, it holds for all natural numbers ( n ).
Step 2
Find the sum of all the natural numbers from 51 to 100, inclusive.
Answer
To calculate the sum from 51 to 100, we can find the sum of the first 100 natural numbers and subtract the sum of the first 50 natural numbers.
The sum of the first 100 natural numbers is: [ S_{100} = \frac{100(101)}{2} = 5050 ]
The sum of the first 50 natural numbers is: [ S_{50} = \frac{50(51)}{2} = 1275 ]
Thus, the sum from 51 to 100 is: [ S = S_{100} - S_{50} = 5050 - 1275 = 3775 ]
Step 3
Given that p = log_x x, express log √x + log(cx) in terms of p.
Answer
Starting with the definitions: [ \log \sqrt{x} = \log x^{1/2} = \frac{1}{2} \log x = \frac{1}{2} p ]
For ( \log(cx) ): Using the product rule, we have: [ \log(cx) = \log c + \log x ] Here, ( \log x = p ): So, [ \log(cx) = \log c + p ]
Thus combining both: [ \log \sqrt{x} + \log(cx) = \frac{1}{2} p + \log c + p = (\frac{3}{2} p + \log c) ]