Problem 258
Question
Sum the series \(1^{2}+\left(1^{2}+2^{2}\right)+\left(1^{2}+2^{2}+3^{2}\right)+\cdots \cdots \cdot\) to \(n\) terms.
Step-by-Step Solution
Verified Answer
The sum of the series up to n terms is given by the formula S = \[ \frac{n(n+1)(n+1)}{4} \]
1Step 1: Recognize the Pattern
The given series is the sum of squares of natural numbers up to a certain term. Specifically, the nth term of the sequence is the sum of the squares of the first n natural numbers.
2Step 2: Apply the Formula
The formula to find the sum of squares of first n natural numbers is \[ \frac{n(n+1)(2n+1)}{6} \] . So, The nth term in the series can be written as Tn = \[ \frac{n(n+1)(2n+1)}{6} \] . Now, the series has become the sum of these terms up to n terms, which is S = T1 + T2 + T3 + ... + Tn.
3Step 3: Find the Series Sum
We need to sum up these Tn terms. So, we have to use the formula for each term and sum them up. The sum up to n terms is given by the following formula S = \[ \frac{1}{6} \sum_{i=1}^{n}i(i+1)(2i+1) \] . Here, \(\sum\) denotes the sum from i=1 to n.
4Step 4: Simplify the Expression
The expression can be simplified further by splitting the triple sum into three separate sums and applying the formulas of each of these sums individually. The sum S thus simplifies to S = \[ \frac{n(n+1)(n+1)}{4} \] .
Key Concepts
Sum of SquaresNatural NumbersSequence Series
Sum of Squares
Understanding the sum of squares is crucial when dealing with sequences involving natural numbers. A sum of squares refers to adding the squares of a series of numbers, like 1, 4, 9, and so on, when these numbers are squared. The mathematical formula to calculate the sum of the first n squares is:\[ \frac{n(n+1)(2n+1)}{6} \]This formula allows you to quickly determine the cumulative sum of squares without manually squaring and adding each number. For example, if you want to find the sum of squares for the first 4 natural numbers, plug in n=4:\[ \frac{4(4+1)(2 \times 4+1)}{6} = 30 \]This expression simplifies to reveal that the sum is 30. This systematic approach makes it easier to handle larger numbers, saving both time and effort.
Natural Numbers
Natural numbers are the set of positive integers starting from 1, 2, 3, and so on. They are the numbers utilized when counting objects. Consequently, natural numbers do not include zero or any negative numbers.
In sequences and series, understanding natural numbers is essential because they form the building blocks of most calculations. For instance, in the given series problem, we use natural numbers to generate terms by squaring each number in a sequence and then progressively adding them up.
With natural numbers, sequences become predictable and calculations manageable. They help in creating seamless formulas for determining sums or patterns within a series.
Sequence Series
A sequence series is a sum of terms arranged in a specific order. In the mathematical context, these terms often follow a certain logic or rule. For instance, in our exercise, we have a sequence where each term involves the sum of squares of natural numbers up to that point.
To comprehend this sequence fully, consider each term as a segment of a larger puzzle. The first term, for instance, includes the square of 1, the second term sums the squares of 1 and 2, and so on, until you reach the nth term. Thus, understanding how each individual part contributes to the entire sequence is crucial.
This process of systematically adding results from each segment illustrates how sequential progressions work, offering deep insights into cumulative calculations. Such series are common in various areas of mathematics and are applied when calculating long sums efficiently.
Other exercises in this chapter
Problem 256
Sum the series \(1 \cdot 2 \cdot 5+2 \cdot 3 \cdot 6+3 \cdot 4 \cdot 7+\cdots \cdots \cdot\) to \(n\) terms.
View solution Problem 257
Sum the series \(1+(1+2)+(1+2+3)+(1+2+3+4)+\cdots \cdots \cdot\) to \(n\) terms.
View solution Problem 259
Sum the series \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\cdots \cdot .\) to 16 terms .
View solution Problem 260
Find the sum of the series \(31^{3}+32^{3}+\cdots \cdots+50^{3}\).
View solution