Stage 1: 1 = 1
Stage 2: 1+3 = 4
Stage 3: 1 +3+5 = 9
Stage 4: 1+3+5+7= 16
a. Find the next four stages and their sums.
b. What is the sum in Stage 20?
c. In what stage will the sum be 567? What is the greatest number added in the sum of this pattern? Explain.
Did you notice that the result is the square of the stage number?
e.g. stage 4 ---->4^2 = 16
carry on
For (b) and (c) you ought to know that the sum of the first n squares is
[n(n+1)(2n+1)]/6
a. To find the next four stages and their sums, you can observe the pattern in the given stages. Notice that in each stage, you start with 1 and then continue adding odd numbers consecutively.
Stage 1: 1
Stage 2: 1 + 3 = 4
Stage 3: 1 + 3 + 5 = 9
Stage 4: 1 + 3 + 5 + 7 = 16
To find the next stage, you add the next odd number. In this case, the next odd number is 9+2=11.
Stage 5: 1 + 3 + 5 + 7 + 11 = 27
To find the sum, you add all the numbers in the stage:
Stage 5 sum: 27
Continuing this pattern:
Stage 6: 1 + 3 + 5 + 7 + 11 + 13 = 40
Stage 7: 1 + 3 + 5 + 7 + 11 + 13 + 15 = 55
Stage 8: 1 + 3 + 5 + 7 + 11 + 13 + 15 + 17 = 72
b. To find the sum in Stage 20, you can continue with the pattern. The 20th stage will have 1 + 3 + 5 + 7 + 11 + 13 + 15 + 17 + 19 + 21 + ... + (2n-1), where n is the stage number.
Using the formula for the sum of an arithmetic sequence:
Sum_n = (n/2)(2a + (n-1)d)
Here, a is the first term (1), n is the stage number (20), and d is the common difference (2). Plugging in the values:
Sum_20 = (20/2)(2(1) + (20-1)(2))
= 10(2 + 19(2))
= 10(2 + 38)
= 10(40)
= 400
So, the sum in Stage 20 is 400.
c. To find the stage where the sum is 567, we can set up the equation:
Sum_n = 567
Again, using the formula for the sum of an arithmetic sequence:
(n/2)(2a + (n-1)d) = 567
Here, a is the first term (1), n represents the stage we are looking for, and d is the common difference (2). Solving this equation can be done through trial and error or using algebraic methods.
To determine the greatest number added in the sum of this pattern, we need to find the last odd number included in the sum at that stage.
In each stage, the last odd number added is represented by (2n - 1), where n is the stage number.
For the sum of 567, we need to find the value of n such that:
(2n - 1) is the largest odd number less than or equal to 567.
We can start by checking values of n by substituting them into the equation:
(2n - 1) ≤ 567
We find that n = 19 satisfies this equation, as (2(19) - 1) = 37 ≤ 567.
Therefore, the sum of 567 will be reached in Stage 19, and the greatest number added in this pattern is (2n - 1) = (2(19) - 1) = 37.