1.(A)(i)Show that 2+4+6+8+..+2n=n(n+1).
(ii)Find the sum of the first 200 even numbers.
(iii)Find the sum of the first 200 odd numbers.
(B)(i)Use the formula at the beginning of the question to find the sum of the first 2n natural numbers.
(ii)Find a formula, in its simplest form, for 1+3+5+7+9+...+(2n-1)
using induction
if n=1, 2 = 1(2) = 2
assuming true for n=k,
check for n=k+1:
2+...+2k+2k+2
= k(k+1) + 2k+2
= k(k+1) + 2(k+1)
= (k+1)(k+2)
as desired.
1+3+5+...+(2n-1) = n^2
(A)(i) To prove that 2+4+6+8+..+2n = n(n+1), we can use mathematical induction.
Step 1: Base case
When n = 1:
2 = 1(1 + 1)
2 = 2
The equation holds true for n = 1.
Step 2: Inductive step
Assume that the formula holds true for some positive integer k, so we have:
2 + 4 + 6 + 8 + ... + 2k = k(k + 1)
Now, we need to prove that the formula also holds true for k + 1.
2 + 4 + 6 + 8 + ... + 2k + 2(k + 1) = (k + 1)(k + 2)
Adding 2(k + 1) to both sides of the equation, we get:
2 + 4 + 6 + 8 + ... + 2k + 2(k + 1) = 2 + 4 + 6 + 8 + ... + 2k + 2k + 2
Using the assumption, we can rewrite the left-hand side as:
k(k + 1) + 2(k + 1)
Factoring out (k + 1), we get:
(k + 1)(k + 2)
Therefore, we have shown that 2 + 4 + 6 + 8 + ... + 2n = n(n + 1) is true for all positive integers n.
(ii) To find the sum of the first 200 even numbers, we can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is given by Sn = (n/2)(a + l), where:
Sn is the sum of the series,
n is the number of terms in the series,
a is the first term, and
l is the last term.
In this case, the first term (a) is 2 and the last term (l) is 2n = 2(200) = 400.
Substituting these values into the formula, we have:
S200 = (200/2)(2 + 400)
S200 = 100(402)
S200 = 40,200
Therefore, the sum of the first 200 even numbers is 40,200.
(iii) To find the sum of the first 200 odd numbers, we can use the formula for the sum of an arithmetic series.
Similar to part (ii), the formula for the sum of an arithmetic series is Sn = (n/2)(a + l), where:
Sn is the sum of the series,
n is the number of terms in the series,
a is the first term, and
l is the last term.
In this case, the first term (a) is 1, and the last term (l) is (2n-1) = 2(200) - 1 = 399.
Substituting these values into the formula, we have:
S200 = (200/2)(1 + 399)
S200 = 100(400)
S200 = 40,000
Therefore, the sum of the first 200 odd numbers is 40,000.
(B)(i) Using the formula at the beginning of the question to find the sum of the first 2n natural numbers, we can let n = n and apply the formula:
Sn = n(n + 1)
Substituting 2n for n, we get:
S2n = (2n)((2n) + 1)
S2n = 2n(2n + 1)
S2n = 4n^2 + 2n
(ii) To find a formula for the sum of the series 1 + 3 + 5 + 7 + 9 + ... + (2n - 1), we can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is given by Sn = (n/2)(a + l), where:
Sn is the sum of the series,
n is the number of terms in the series,
a is the first term, and
l is the last term.
In this case, the first term (a) is 1 and the last term (l) is (2n - 1).
Substituting these values into the formula, we have:
Sn = (n/2)(1 + (2n - 1))
Sn = (n/2)(2n)
Sn = n^2
Therefore, the simplified formula for 1 + 3 + 5 + 7 + 9 + ... + (2n - 1) is n^2.
To prove the formula for the sum of even numbers and odd numbers, we can use mathematical induction.
(A)(i) To prove that 2 + 4 + 6 + 8 + ... + 2n = n(n + 1), we need to follow these steps:
Step 1: Base case:
For n = 1, the left side of the equation becomes 2, and the right side becomes 1(1 + 1) = 2. The equation holds for the base case.
Step 2: Inductive step:
Assume the equation holds for some k, i.e., 2 + 4 + 6 + ... + 2k = k(k + 1).
Now, we need to show that the equation holds for k + 1, i.e., 2 + 4 + 6 + ... + 2(k + 1) = (k + 1)(k + 2).
Starting with the left side of the equation:
2 + 4 + 6 + ... + 2(k + 1) = [2 + 4 + 6 + ... + 2k] + 2(k + 1)
= k(k + 1) + 2(k + 1) (using the assumption from the inductive step)
= (k + 1)(k + 2).
Therefore, the equation holds for k + 1 as well.
By proving the base case and the inductive step, we can conclude that 2 + 4 + 6 + ... + 2n = n(n + 1).
(ii) The sum of the first 200 even numbers can be found using the formula derived above.
For n = 100 (since there are 200 even numbers), we have:
2 + 4 + 6 + ... + 2(100) = 100(100 + 1)
= 100(101)
= 10,100.
Therefore, the sum of the first 200 even numbers is 10,100.
(iii) The sum of the first 200 odd numbers can also be calculated using a similar method.
For n = 100 (since there are 200 odd numbers), we have:
1 + 3 + 5 + ... + (2(100) - 1) = 100(100)
= 10,000.
Therefore, the sum of the first 200 odd numbers is 10,000.
(B)(i) To find the sum of the first 2n natural numbers, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(first term + last term), where the first term is 1 and the last term is 2n.
Substituting the values into the formula:
Sum = (n/2)(1 + 2n)
= (n/2 + n^2).
Therefore, the sum of the first 2n natural numbers is n/2 + n^2.
(ii) To find a formula for the sum of the series 1 + 3 + 5 + 7 + 9 + ... + (2n - 1), we can observe that each term is an odd number and increases by 2.
The sum of the series will be the sum of all the odd numbers starting from 1 up to (2n - 1).
Using the formula for the sum of an arithmetic series, the sum of the first n odd numbers is given by:
Sum = (n/2)(first term + last term)
= (n/2)(1 + (2n - 1))
= (n/2)(2n)
= n^2.
Therefore, the sum of the series 1 + 3 + 5 + 7 + 9 + ... + (2n - 1) is n^2.