Use the method of differences to prove the following formulas
1) ∑_(r=1)^n▒〖r=(n(n+1))/2〗
2) ∑_(r=1)^n▒〖r^2=(n(n+1)(2n+1))/6〗
To prove the formulas using the method of differences, we start by calculating the differences between consecutive terms of the given series. Let's begin with the first formula:
1) ∑_(r=1)^n▒r = (n(n+1))/2
First, calculate the differences between consecutive terms:
2 - 1, 3 - 2, 4 - 3, ..., n - (n-1)
We can observe a pattern in these differences. The first difference is always 1, the second difference is always 1, and so on. Therefore, for any series of n terms, we will have n - 1 differences, all equal to 1.
Now, let's calculate the differences between the generated differences:
1 - 1, 1 - 1, 1 - 1, ..., 1 - 1
Here, we can see that the second differences are all equal to 0. This means that the original series follows a quadratic pattern, as the differences become constant.
Using this information, we can assume that the sum of consecutive integers can be represented by a quadratic equation of the form:
an^2 + bn + c
To find the coefficients a, b, and c, we can use the initial terms of the given series:
When n = 1, the sum is 1.
When n = 2, the sum is 3.
When n = 3, the sum is 6.
Setting up a system of equations using these initial terms:
a + b + c = 1 (equation 1)
4a + 2b + c = 3 (equation 2)
9a + 3b + c = 6 (equation 3)
Solving this system of equations, we find that a = 1/2, b = 1/2, and c = 0.
Therefore, the sum of the series ∑_(r=1)^n▒r is given by:
(n(n+1))/2 = (1/2)n^2 + (1/2)n + 0
= (n(n+1))/2
Thus, we have proved the first formula using the method of differences.
Now let's move on to the second formula:
2) ∑_(r=1)^n▒r^2 = (n(n+1)(2n+1))/6
First, calculate the differences between consecutive terms:
(2^2 - 1^2), (3^2 - 2^2), (4^2 - 3^2), ..., (n^2 - (n-1)^2)
We can simplify these differences:
(4 - 1), (9 - 4), (16 - 9), ..., (n^2 - (n-1)^2)
This simplifies further to:
3, 5, 7, ..., (2n - 1)
We can observe a pattern in the differences. The first difference is 2, the second difference is also 2, and so on. Therefore, for any series of n terms, we will have n - 1 differences, all equal to 2.
Now, let's calculate the differences between the generated differences:
2 - 2, 2 - 2, 2 - 2, ..., 2 - 2
Here, we can see that the second differences are all equal to 0. This means that the original series follows a cubic pattern, as the differences become constant.
Using this information, we can assume that the sum of consecutive squares can be represented by a cubic equation of the form:
an^3 + bn^2 + cn + d
To find the coefficients a, b, c, and d, we can use the initial terms of the given series:
When n = 1, the sum is 1.
When n = 2, the sum is 5.
When n = 3, the sum is 14.
Setting up a system of equations using these initial terms:
a + b + c + d = 1 (equation 1)
8a + 4b + 2c + d = 5 (equation 2)
27a + 9b + 3c + d = 14 (equation 3)
Solving this system of equations, we find that a = 1/3, b = 1/2, c = 1/6, and d = 0.
Therefore, the sum of the series ∑_(r=1)^n▒r^2 is given by:
(n(n+1)(2n+1))/6 = (1/3)n^3 + (1/2)n^2 + (1/6)n + 0
= (n(n+1)(2n+1))/6
Thus, we have proved the second formula using the method of differences.