Question

The sum to \(n\) terms of the series \(1^{2}+\left(1^{2}+3^{2}\right)+\left(1^{2}+3^{2}+5^{2}\right)+\ldots\) is : (a) \(\frac{1}{3}\left(n^{4}+2 n^{2}\right)\) (b) \(\frac{1}{3}\left(n^{3}+3 n^{2}-n\right)\) (c) \(\frac{1}{6} n(n+1)\left(2 n^{2}+2 n-1\right)\) (d) none of these

Short answer

Answer: The closed-form expression for the sum of the series up to its nth term is \(S_n = \frac{1}{6} n(n+1)\left(2 n^{2}+2 n-1\right)\).

Step-by-step solution

Identify the pattern of the series

The series given is: \(1^{2}+\left(1^{2}+3^{2}\right)+\left(1^{2}+3^{2}+5^{2}\right)+\ldots\) Notice that the terms in each parenthesis are all consecutive odd numbers, and they form a cumulative sum of squares of each odd number until the current term.

Rewrite the series in the sigma notation

We can express the nth term of the series as a sum of squares of the first n odd numbers, i.e., \(S_n = \sum_{k=1}^n (2k-1)^2\). So we need to find the closed-form expression for this sum.

Simplify the term inside the summation

Let's simplify \((2k-1)^2\) and rewrite it as a polynomial in k: \((2k-1)^2 = 4k^2 - 4k + 1\)

Find the closed-form expression for the sum

Now, we need to find the sum of this polynomial for all terms from k=1 to k=n: \(S_n = \sum_{k=1}^n (4k^2 - 4k + 1)\) We can split the summation and use well-known summation formulas to find the closed-form expression: \(S_n = 4\sum_{k=1}^n k^2 - 4\sum_{k=1}^n k + \sum_{k=1}^n 1\) Using the formulas for the sum of squares, sum of integers, and sum of constants, we get: \(S_n = 4\frac{n(n+1)(2n+1)}{6} - 4\frac{n(n+1)}{2} + n\) \(S_n = \frac{2}{3}n(n+1)(2n+1) - 2n(n+1) + n\)

Simplify the expression and match with the given options

Simplify the expression for \(S_n\): \(S_n = n(n+1)\left(\frac{2}{3}(2n+1) - 2\right) + n\) \(S_n = n(n+1)\left(\frac{4n+2}{3} - \frac{6}{3}\right) + n\) \(S_n = n(n+1)\left(\frac{4n-4}{3}\right) + n\) \(S_n = \frac{1}{3}n(n+1)(4n-4) + n\) Now compare the final expression of \(S_n\) with the given options. It matches with option (c), so the correct answer is: (c) \(\frac{1}{6} n(n+1)\left(2 n^{2}+2 n-1\right)\)

Key concepts

Sigma Notation

Understanding sigma notation is crucial for summing sequences and series. Sigma (\(\Sigma\)) notation is a compact way to express the sum of a sequence. Instead of writing out long sums, you can use this notation to simplify your expressions.

• The symbol \(\Sigma\) stands for "sum", and it is followed by an expression that tells you what to sum and the range of the summation.
• The notation \(\sum_{k=1}^{n} a_k\) means to sum up all the terms \(a_k\) where \(k\) starts at 1 and goes up to \(n\).

In this series, each term involves summing squares of consecutive odd numbers, which can efficiently be expressed using sigma notation. Writing this as \(\sum_{k=1}^{n} (2k-1)^2\) allows us to easily manipulate and find closed forms.

Sum of Odd Numbers

Odd numbers are numbers that are not divisible by 2, and they follow a regular pattern: 1, 3, 5, 7, etc. This sequence can also be expressed as \(2k-1\) for \(k = 1, 2, 3, \ldots\). When adding odd numbers, especially their squares, there's a neat pattern or formula that emerges:

• For instance, the sum of the first \(n\) odd numbers is equal to \(n^2\).
• Furthermore, the series \(1^2 + 3^2 + 5^2 + \ldots + (2n-1)^2\) amounts to the cumulative sum of squares of the first \(n\) odd numbers.

Recognizing and working with these patterns simplifies finding sums in series problems.

Closed-form Expression

Closed-form expressions are formulas that provide a direct way to calculate a sum without going through iterative or recursive sums. Instead of adding elements one by one, you can plug values into a formula.

• For example, the sum of the first \(n\) squares \(\sum_{k=1}^{n} k^2\) is given by the closed expression \(\frac{n(n+1)(2n+1)}{6}\).
• This helps in computational efficiency and makes it easier to handle series with a large number of terms.

By finding the closed-form expression in problems, we can determine the sum quickly. This is especially useful in the series problems where a pattern repeats systematically.

Sum of Squares

The sum of squares is a specific type of series where each term is squared. It has special significance and applications in mathematics, particularly in areas of statistical variance and geometry.

• The formula for the sum of squares \(\sum_{k=1}^n k^2\) is \(\frac{n(n+1)(2n+1)}{6}\).
• This formula allows for quick calculations without explicit summation.
• In series problems, such as the one provided, understanding how to express odd numbers as squares simplifies finding solutions.

When combined with sigma notation and closed-form techniques, the sum of squares can be calculated swiftly. This understanding is vital for efficiently solving series—and other cumulative problems.