Sum of Squares of Positive Integers Less Than or Equal to 100

In mathematics, the sum of squares of positive integers can be an interesting and important topic for various fields, including number theory, algebra, and computer science. This article focuses on the sum of squares of all positive integers less than or equal to 100, providing a detailed breakdown of the calculation process and the underlying mathematical principles.

Understanding the Problem

The problem at hand is to find the sum of the squares of all positive integers from 1 to 100. In mathematical terms, we are looking for the value of the following expression:

12 22 32 ... 1002

Step-by-Step Calculation

Let's break down the calculation step by step:

1. Notation and Initial Setup

We start by defining the general term of the sequence as n2, where n is a positive integer ranging from 1 to 100.

2. First Difference Calculation

To find the first difference, we calculate the difference between consecutive terms:

First Difference: n2 - (n-1)2 n2 - (n2 - 2n 1) 2n - 1

3. Second Difference Calculation

Next, we find the second difference by calculating the difference between consecutive first differences:

Second Difference: (2n - 1) - (2(n-1) - 1) (2n - 1) - (2n - 3) 2

4. Third Difference Calculation

Finally, we calculate the third difference (which is a constant in this case):

Third Difference: 2 - 2 0

Formulating a General Equation

The second difference being constant suggests that the sequence of partial sums can be expressed as a cubic polynomial. Let's derive the general term of this polynomial:

5. Formulating the Polynomial

Assuming the general term is of the form:

tn an3 bn2 cn d

6. Solving for the Coefficients

We need to find the coefficients a, b, c, and d. Using the given values:

For n 1: a(1)3 b(1)2 c(1) d 1

8a(4)3 4b(4)2 2c(4) d 5

27a(9)3 9b(9)2 3c(9) d 14

64a(16)3 16b(16)2 4c(16) d 30

Solving this system of equations, we find:

a 1/3, b 1/2, c 1/6, and d 0

7. Substituting the Coefficients

The general term of the sequence is now:

tn (1/3)n3 (1/2)n2 (1/6)n

Applying the General Term to n 100

To find the sum of the squares of the first 100 positive integers, we substitute n 100 into the general term:

t100 (1/3)(100)3 (1/2)(100)2 (1/6)(100)

t100 (1/3)(1000000) (1/2)(10000) (50)

t100 333333.333 5000 50

t100 338350

Therefore, the sum of the squares of all positive integers less than or equal to 100 is 338350.

Conclusion

This detailed calculation highlights the power and utility of polynomial equations in solving complex mathematical problems. Understanding these concepts allows us to compute the sum of squares for any range of positive integers, which can be invaluable in various applications.