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(a) (i) Prove by induction that, for any n, the sum of the first n natural numbers is \( rac{n(n+1)}{2}\) - Leaving Cert Mathematics - Question 2 - 2014
Question 2
(a) (i) Prove by induction that, for any n, the sum of the first n natural numbers is \( rac{n(n+1)}{2}\). (ii) Find the sum of all the natural numbers from 51 to 1... 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 \( rac{n(n+1)}{2}\) - Leaving Cert Mathematics - Question 2 - 2014
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 by induction, we follow these steps:
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Base Case: Check for (n = 1):
The sum of the first natural number is (1). According to the formula:
[ \frac{1(1+1)}{2} = 1 ]
Therefore, the statement holds for (n = 1). -
Induction Hypothesis: Assume true for (n = k):
We assume that the sum of the first (k) natural numbers is (\frac{k(k+1)}{2}).
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Induction Step: Show true for (n = k + 1):
The sum of the first (k + 1) natural numbers is:
[ 1 + 2 + ... + k + (k + 1) = \frac{k(k+1)}{2} + (k + 1) ]Combining both parts:
[ \frac{k(k+1)}{2} + (k + 1) = \frac{k(k+1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2} ]Thus, the formula holds for (n = k + 1).
Since both the base case and the induction step are proven, by induction, the formula 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 from 51 to 100, we can use the formula for the sum of the first n natural numbers.
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Calculate the sum from 1 to 100:
[ \text{Sum from 1 to 100} = \frac{100(100 + 1)}{2} = 5050 ] -
Calculate the sum from 1 to 50:
[ \text{Sum from 1 to 50} = \frac{50(50 + 1)}{2} = 1275 ] -
Subtract the two sums to find the sum from 51 to 100:
[ 5050 - 1275 = 3775 ]
Therefore, the sum of all natural numbers from 51 to 100 is 3775.
Step 3
Given that p = log_x, express log_c(√x) + log_c(acx) in terms of p.
Answer
Using the properties of logarithms, we can express ( \log_c(\sqrt{x}) + \log_c(acx) ) as follows:
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Logarithm of Square Root:
[ \log_c(\sqrt{x}) = \log_c(x^{1/2}) = \frac{1}{2} \log_c(x) ] -
Logarithm of Product:
[ \log_c(acx) = \log_c(a) + \log_c(c) + \log_c(x) ]Simplifying further, since ( c = 1 ):
[ \log_c(acx) = \log_c(a) + p ] -
Combine the two results:
[ \log_c(\sqrt{x}) + \log_c(acx) = \frac{1}{2}p + \log_c(a) + p = \left(\frac{3}{2}p + \log_c(a)\right) ]
Thus, the expression in terms of p is ( \frac{1}{2}p + \log_c(a) + p ).