Special Series

Here we are going to look at some special series which are very useful in finding the sum of various series.

Sum of first n natural numbers

\[1+2+3+...+n = \frac{n(n+1)}{2}\]

The sum of the first \(n\) natural numbers can be derived from the fact that these numbers form an Arithmetic Progression (AP) with the first term \(a_1 = 1\), the common difference \(d = 1\), and the last term \(a_n = n\). The sum of an AP can be calculated using the formula:

\[ S_n = \frac{n}{2}(a_1 + a_n) \]

For the series of the first \(n\) natural numbers, this becomes:

\[ S_n = \frac{n}{2}(1 + n) \]

Here, \(n\) represents the number of terms, \(1\) is the first term, and \(n\) is the last term of the sequence. Substituting these values into the formula gives us:

\[ S_n = \frac{n(n + 1)}{2} \]

This formula accounts for the linear increase of natural numbers and utilizes the characteristic of arithmetic progressions that the average of the first and last term, multiplied by the number of terms, gives the total sum.

Sum of square of first n natural numbers

The sum of the squares of the first \(n\) natural numbers is given by the formula:

\[ 1^2+2^2+3^2+...+n^2 = \frac{n(n+1)(2n+1)}{6} \]

Proof:

To prove this theorem, we utilize the identity:

\[ (k+1)^3 - k^3 = 3k^2 + 3k + 1 \]

This identity represents the difference between the cubes of consecutive natural numbers and can be derived from the binomial expansion of \((k+1)^3\).

Step 1: Form \(n\) equations by putting \(k = 1, 2, 3, \ldots, n\) into the identity:

For \(k = 1\): \(2^3 - 1^3 = 3 \cdot 1^2 + 3 \cdot 1 + 1\)

For \(k = 2\): \(3^3 - 2^3 = 3 \cdot 2^2 + 3 \cdot 2 + 1\)

\(\vdots\)

For \(k = n\): \((n+1)^3 - n^3 = 3n^2 + 3n + 1\)

Step 2: Add all \(n\) equations together:

\[ \sum_{k=1}^{n} [(k+1)^3 - k^3] = \sum_{k=1}^{n} (3k^2 + 3k + 1) \]

The left side telescopes, meaning most terms cancel out, leaving only the first and last term:

\[ (n+1)^3 - 1 = 3 \sum_{k=1}^{n} k^2 + 3 \sum_{k=1}^{n} k + n \]

We know that \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\) from the formula for the sum of the first \(n\) natural numbers.

Substituting this into our equation gives:

\[ n^3 + 3n^2 + 3n + 1 - 1 = 3 \sum_{k=1}^{n} k^2 + 3 \cdot \frac{n(n+1)}{2} + n \]

Simplifying further:

\[ n^3 + 3n^2 + 3n = 3 \sum_{k=1}^{n} k^2 + \frac{3n^2}{2} + \frac{3n}{2} + n \]

Rearranging for \(\sum_{k=1}^{n} k^2\):

\[ 3 \sum_{k=1}^{n} k^2 = n^3 + 3n^2 + 3n - \frac{3n^2}{2} - \frac{3n}{2} - n \]

\[ 3 \sum_{k=1}^{n} k^2 = n^3 + \frac{3n^2}{2} + \frac{n}{2} \]

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

This completes the proof, demonstrating that the sum of the squares of the first \(n\) natural numbers is \(\frac{n(n+1)(2n+1)}{6}\).

Sum of cubes of first n natural numbers

The sum of the cubes of the first \(n\) natural numbers is given by the formula:

\[ 1^3+2^3+3^3+\ldots+n^3 = \left( \frac{n(n+1)}{2} \right)^2 = \left(1+2+3+\ldots+n\right)^2\]

Proof:

To prove this identity, we use the identity:

\[ (k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1 \]

This identity represents the difference between the fourth powers of consecutive natural numbers, and can be derived from the binomial theorem.

Step 1: Form \(n\) equations by substituting \(k = 1, 2, 3, \ldots, n\) into the identity:

For \(k = 1\): \(2^4 - 1^4 = 4 \cdot 1^3 + 6 \cdot 1^2 + 4 \cdot 1 + 1\)

For \(k = 2\): \(3^4 - 2^4 = 4 \cdot 2^3 + 6 \cdot 2^2 + 4 \cdot 2 + 1\)

\(\vdots\)

For \(k = n\): \((n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1\)

Step 2: Add all \(n\) equations together:

The left side telescopes, meaning most terms cancel out, leaving only the first and last term:

\[ (n+1)^4 - 1 = 4 \sum_{k=1}^{n} k^3 + 6 \sum_{k=1}^{n} k^2 + 4 \sum_{k=1}^{n} k + n \]

We already know the formulas for the sums of the first \(n\) natural numbers and their squares:

• \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
• \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)

Substituting these known formulas:

\[ (n+1)^4 - 1 = 4 \sum_{k=1}^{n} k^3 + 6 \left( \frac{n(n+1)(2n+1)}{6} \right) + 4 \left( \frac{n(n+1)}{2} \right) + n \]

Simplifying, we aim to solve for \(\sum_{k=1}^{n} k^3\):

\[ 4 \sum_{k=1}^{n} k^3 = (n+1)^4 - 1 - n(n+1)(2n+1) - 2n(n+1) - n \]

Further simplification yields the formula for the sum of the cubes:

\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \]

This completes the proof, showing that the sum of the cubes of the first \(n\) natural numbers is indeed the square of the sum of the first \(n\) natural numbers.

Applications

Example

Problem: Find the sum of the series \(1^2 + 3^2 + 5^2 + \ldots\) up to \(n\) terms, where each term is the square of an odd number.

Solution:

Step 1: Express the series in sigma notation. The \(n\)-th term of the series, being the square of the \(n\)-th odd number, can be expressed as \((2n - 1)^2\). Thus, the series sum \(S\) can be written as:

\[ S = \sum_{k=1}^{n} (2k-1)^2 \]

Step 2: Expand the polynomial inside the sigma notation.

\[ S = \sum_{k=1}^{n} (4k^2 - 4k + 1) \]

Step 3: Distribute the sigma across the terms.

\[ S = 4\sum_{k=1}^{n} k^2 - 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

Step 4: Apply special series formulas. The sum of the first \(n\) squares is \(\frac{n(n + 1)(2n + 1)}{6}\), the sum of the first \(n\) natural numbers is \(\frac{n(n + 1)}{2}\), and the sum of \(n\) ones is \(n\).

\[ S = 4\left(\frac{n(n + 1)(2n + 1)}{6}\right) - 4\left(\frac{n(n + 1)}{2}\right) + n \]

Step 5: Simplify the expression.

\[ S = \frac{2n(n + 1)(2n + 1)}{3} - 2n(n + 1) + n \]

\[ = \frac{2n(2n^2 + 3n + 1) - 6n(n + 1) + 3n}{3} \]

\[ = \frac{4n^3 + 6n^2 + 2n - 6n^2 - 6n + 3n}{3} \]

\[ = \frac{4n^3 - n}{3} \]

Conclusion:

\[ S = \frac{4n^3 - n}{3} \]

Therefore, the sum of the series \(1^2 + 3^2 + 5^2 + \ldots\) up to \(n\) terms is \(\frac{4n^3 - n}{3}\).

Example

Problem: Find the sum of the series \(1 \cdot 1^2 + 2 \cdot 3^2 + 3 \cdot 5^2 + 4 \cdot 7^2 + \ldots\) up to \(n\) terms.

Solution:

Step 1: Identify the pattern in the series. The series is composed of terms in which the coefficients (1, 2, 3, 4, ...) increase linearly and the squares are of the sequence of odd numbers (1, 3, 5, 7, ...), which form an arithmetic progression (AP) with a common difference of 2.

Step 2: Express the \(j\)-th term of the series using the AP property. The \(j\)-th odd number can be represented as \(2j - 1\). Thus, the \(j\)-th term of the series can be written as:

\[ T_j = j(2j - 1)^2 \]

Step 3: Write the series in summation form to represent the sum \(S\) of \(n\) terms:

\[ S = \sum_{j=1}^{n} j(2j - 1)^2 \]

Step 4: Expand the polynomial inside the sigma notation:

\[ S = \sum_{j=1}^{n} j(4j^2 - 4j + 1) \]

\[ S = \sum_{j=1}^{n} (4j^3 - 4j^2 + j) \]

Step 5: Distribute the sigma across the terms:

\[ S = 4\sum_{j=1}^{n} j^3 - 4\sum_{j=1}^{n} j^2 + \sum_{j=1}^{n} j \]

Step 6: Apply special series formulas. The sum of the first \(n\) cubes is \(\frac{n^2(n + 1)^2}{4}\), the sum of the first \(n\) squares is \(\frac{n(n + 1)(2n + 1)}{6}\), and the sum of the first \(n\) natural numbers is \(\frac{n(n + 1)}{2}\).

\[ S = 4\left(\frac{n^2(n + 1)^2}{4}\right) - 4\left(\frac{n(n + 1)(2n + 1)}{6}\right) + \frac{n(n + 1)}{2} \]

Step 7: Simplify the expression:

\[ \begin{align*} S & = n^2(n + 1)^2 - \frac{2n(n + 1)(2n + 1)}{3} + \frac{n(n + 1)}{2} \\ & = n(n+1) \left(n(n + 1) - \frac{2(2n + 1)}{3} + \frac{1}{2}\right) \\ & = n(n+1) \left(\frac{6n(n + 1) - 4(2n + 1) + 3}{6}\right) \\ & = n(n+1) \left(\frac{6n^2 + 6n - 8n - 4 + 3}{6}\right) \\ & = n(n+1) \left(\frac{6n^2 - 2n - 1}{6}\right) \\ & \implies S = \frac{n(n+1)(6n^2 - 2n - 1)}{6}. \end{align*} \]

Example

Problem: Find the sum of the series \(1 + (2+3) + (3+4+5) + (4+5+6+7) + \ldots\) up to \(n\) groups.

Solution:

Step 1: Observe the Pattern of Groups
Each group starts with consecutive integers increasing by 1 from each previous group's starting number. The first term of the \(k\)-th group can be observed as starting from \(k\).

Step 2: Determine the Number of Terms in the \(k\)-th Group
The \(k\)-th group contains \(k\) terms, starting from \(k\).

Step 3: Sum Each Group Using the Arithmetic Progression Formula
Since the terms in each group form an arithmetic progression (AP) with a common difference of 1, the sum of the \(k\)-th group, \(S_k\), can be found using the formula for the sum of an AP:

\[ S_k = \frac{n}{2} [2a + (n-1)d] \]

where \(n\) is the number of terms, \(a\) is the first term, and \(d\) is the common difference. For the \(k\)-th group:

\[ n = k, \; a = k, \; d = 1 \]

So, the sum of the \(k\)-th group is:

\[ S_k = \frac{k}{2} [2k + (k-1)\cdot1] = \frac{k}{2} (2k + k - 1) = \frac{k}{2} (3k - 1) \]

Step 4: Write the Series Sum in Sigma Notation and Sum
The total sum of the series up to \(n\) groups is:

\[ S = \sum_{k=1}^{n} S_k = \sum_{k=1}^{n} \frac{k(3k - 1)}{2} \]

Step 5: Simplify the Sigma Expression
To simplify, we distribute the sigma:

\[ S = \frac{1}{2} \sum_{k=1}^{n} (3k^2 - k) \]

\[ S = \frac{1}{2} \left( 3\sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k \right) \]

Apply the formulas for the sum of the first \(n\) squares and the sum of the first \(n\) natural numbers:

\[ S = \frac{1}{2} \left( 3\frac{n(n + 1)(2n + 1)}{6} - \frac{n(n + 1)}{2} \right) \]

\[ S = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1)}{2} - \frac{n(n + 1)}{2} \right) \]

\[ S = \frac{n(n + 1)}{4} \left( (2n + 1) - 1 \right) \]

\[ S = \frac{n(n + 1)(2n)}{4} \]

\[ S = \frac{n^2(n + 1)}{2} \]

Conclusion:
The sum of the series \(1 + (2+3) + (3+4+5) + (4+5+6+7) + \ldots\) up to \(n\) groups is \(\frac{n^2(n + 1)}{2}\).

Example

Problem: Find the sum of the series \(1 + (2+3) + (4+5+6) + (7+8+9+10) + \ldots\) up to \(n\) groups.

Solution:

Step 1: Identify the Pattern
This series is composed of groups where each group starts with the next consecutive number following the last number of the previous group, and each group contains one more term than the previous group. The first group has one term, the second group has two terms, and so on.

Step 2: Determine the First Term of Each Group
The series of first terms for each group forms a sequence that does not follow a simple arithmetic progression due to the increasing number of terms in each group. However, we can observe the first term of the \(k\)-th group is equal to \([1+2+3+...+(k-1)]+1\)\(=\frac{(k-1)k}{2} + 1\).

Step 3: Sum Each Group
Each group forms an arithmetic progression with a common difference of 1. The sum of the \(k\)-th group, \(S_k\), where the first term is \(a_k = \frac{(k-1)k}{2} + 1\) and the number of terms is \(k\), can be found using the formula for the sum of an AP:

\[ S_k = \frac{k}{2} [2a_k + (k-1)d] \]

Since \(d = 1\), we simplify to:

\[ S_k = \frac{k}{2} [2\left(\frac{(k-1)k}{2} + 1\right) + (k-1)] \]

\[ S_k = \frac{k}{2} \left(k^2 - k + 2 + k - 1\right) \]

\[ S_k = \frac{k}{2} (k^2 + 1) \]

\[ S_k = \frac{k(k^2 + 1)}{2} \]

Step 4: Sum the Series up to \(n\) Groups
The total sum \(S\) of the series up to \(n\) groups is:

\[ S = \sum_{k=1}^{n} \frac{k(k^2 + 1)}{2} \]

\[ S = \frac{1}{2} \sum_{k=1}^{n} (k^3 + k) \]

Using the sum of cubes and the sum of natural numbers formulas, we get:

\[ S = \frac{1}{2} \left( \sum_{k=1}^{n} k^3 + \sum_{k=1}^{n} k \right) \]

\[ S = \frac{1}{2} \left( \left(\frac{n(n + 1)}{2}\right)^2 + \frac{n(n + 1)}{2} \right) \]

\[ S = \frac{1}{2} \left( \frac{n^2(n + 1)^2}{4} + \frac{n(n + 1)}{2} \right) \]

\[ S = \frac{n(n + 1)}{4} \left( \frac{n(n + 1)}{2} + 1 \right) \]

\[ S = \frac{n(n + 1)}{4} \left( \frac{n^2 + n + 2}{2} \right) \]

\[ S = \frac{n(n + 1)(n^2 + n + 2)}{8} \]

Conclusion:
The sum of the series \(1 + (2+3) + (4+5+6) + (7+8+9+10) + \ldots\) up to \(n\) groups is \(\frac{n(n + 1)(n^2 + n + 2)}{8}\).

Example

Problem: Find the sum of the series \(11^2 + 12^2 + 13^2 + \ldots + 20^2\).

Solution 1:Solution 2:

Step 1: Express the Series in Summation Form
The given series can be initially written in summation form as \( \sum_{r=11}^{20} r^2 \). However, to simplify the calculation, we aim to adjust the indices so that \(r\) starts from 1.

Step 2: Adjust the Summation to Start from 1
To adjust the summation index to start from 1 instead of 11, we note that each term in the series can be expressed as \((10 + k)^2\) where \(k\) ranges from 1 to 10. Thus, the series becomes:

\[ \sum_{k=1}^{10} (10+k)^2 \]

Step 3: Expand the Series
Expanding the quadratic expression inside the summation gives us:

\[ \sum_{k=1}^{10} (100 + 20k + k^2) \]

\[ = \sum_{k=1}^{10} 100 + \sum_{k=1}^{10} 20k + \sum_{k=1}^{10} k^2 \]

Step 4: Simplify the Summation
We can now calculate each part of the summation separately:

1. \( \sum_{k=1}^{10} 100 = 100 \times 10 \)
2. \( \sum_{k=1}^{10} 20k = 20 \sum_{k=1}^{10} k \)
3. \( \sum_{k=1}^{10} k^2 \) can be calculated using the formula for the sum of the squares of the first \(n\) natural numbers, \( \frac{n(n + 1)(2n + 1)}{6} \).

Step 5: Calculate the Sums

1. \( \sum_{k=1}^{10} 100 = 1000 \)
2. \( \sum_{k=1}^{10} 20k = 20 \times \frac{10(10 + 1)}{2} = 20 \times 55 = 1100 \)
3. \( \sum_{k=1}^{10} k^2 = \frac{10(10 + 1)(2 \times 10 + 1)}{6} = \frac{10 \times 11 \times 21}{6} = 385 \)

Step 6: Combine the Results
Combining all parts gives the total sum:

\[ S = 1000 + 1100 + 385 = 2485 \]

Conclusion:
The sum of the series \(11^2 + 12^2 + 13^2 + \ldots + 20^2\) is 2485.

Step 1: Write the Desired Sum in Terms of Known Series
The sum of squares from 11 to 20 can be represented as the difference between the sum of the first 20 squares and the first 10 squares:

\[ S = \sum_{r=1}^{20} r^2 - \sum_{r=1}^{10} r^2 \]

Step 2: Apply the Formula for the Sum of Squares
The formula for the sum of the first \(n\) squares is \(\frac{n(n + 1)(2n + 1)}{6}\). We apply this formula to both parts of the equation:

\[ S = \frac{20(20 + 1)(2 \times 20 + 1)}{6} - \frac{10(10 + 1)(2 \times 10 + 1)}{6} \]

Step 3: Calculate Each Sum Individually

• For the first 20 squares:

\[ \frac{20(21)(41)}{6} = \frac{20 \times 21 \times 41}{6} \]

• For the first 10 squares:

  \[ \frac{10(11)(21)}{6} = \frac{10 \times 11 \times 21}{6} \]

Therefore,

\[ S = \frac{20 \times 21 \times 41}{6} - \frac{10 \times 11 \times 21}{6} \]

\[ S = 2870 - 385 = 2485 \]

Conclusion:
The sum of the series \(11^2 + 12^2 + 13^2 + \ldots + 20^2\) is 2485, as confirmed by this alternative method.

Dealing with alternating Series

Example

Problem: Find the sum of the series \(1^2 - 2^2 + 3^2 - 4^2 + \ldots\) up to \(2n\) terms.

Solution 1: By Segregating in to two seriesSolution 2: By pairing consecutive terms

Step 1: Segregate the Positive and Negative Terms
The series alternates between positive and negative squares. We can separate these into two series: one for the positive terms and one for the negative terms.

1. Positive terms: \(1^2 + 3^2 + 5^2 + \ldots\) up to \(n\) terms.
2. Negative terms: \(- (2^2 + 4^2 + 6^2 + \ldots)\) up to \(n\) terms.

Step 2: Express Each Series in Summation Form

1. Positive series: \(\sum_{k=1}^{n} (2k-1)^2\), since the positive terms are the squares of odd numbers.
2. Negative series: \(-\sum_{k=1}^{n} (2k)^2\), since the negative terms are the squares of even numbers.

Step 3: Calculate the Sum

1. Positive series sum:

   \[ \sum_{k=1}^{n} (2k-1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) \]
   \[ = 4\sum_{k=1}^{n} k^2 - 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]
   \[ = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} - 4 \cdot \frac{n(n + 1)}{2} + n \]

2. Negative series sum:

   \[ -\sum_{k=1}^{n} (2k)^2 = -4\sum_{k=1}^{n} k^2 \]
   \[ = -4 \cdot \frac{n(n + 1)(2n + 1)}{6} \]

Step 5: Combine the Sums
Combine the positive and negative series sums to get the total sum:

\[ S = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} - 4 \cdot \frac{n(n + 1)}{2} + n - 4 \cdot \frac{n(n + 1)(2n + 1)}{6} \]

\[ = -4 \cdot \frac{n(n + 1)}{2} + n \]

\[ = -2n(n + 1) + n \]

\[ = n - 2n^2 - 2n \]

\[ = -2n^2 - n \]

Conclusion:
The sum of the series \(1^2 - 2^2 + 3^2 - 4^2 + \ldots\) up to \(2n\) terms is \(-2n^2 - n\).

Problem: Find the sum of the series \(1^2 - 2^2 + 3^2 - 4^2 + \ldots\) up to \(2n\) terms by pairing consecutive terms to form \(n\) pairs.

Solution:

Step 1: Pair Consecutive Terms
Pair the terms as follows: \((1^2 - 2^2) + (3^2 - 4^2) + \ldots + ([(2k-1)^2 - (2k)^2] + \ldots + [(2n-1)^2 - 2n^2])\), creating \(n\) pairs.

Step 2: Find the General Term for the \(k\)-th Pair
The general term for the \(k\)-th pair can be calculated as:

\[ (2k-1)^2 - (2k)^2 \]

Step 3: Simplify the General Term
Simplify the expression within the \(k\)-th pair:

\[ = (4k^2 - 4k + 1) - (4k^2) \]

\[ = -4k + 1 \]

Step 4: Write the Sum Using Sigma Notation and Simplify
Express the sum of the series as the sum of \(n\) pairs:

\[ S = \sum_{k=1}^{n} (-4k + 1) \]

Simplify the sum by distributing the sigma notation:

\[ S = -4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

Apply the formula for the sum of the first \(n\) natural numbers, \(\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}\), and the sum of \(n\) ones, \(\sum_{k=1}^{n} 1 = n\):

\[ S = -4 \cdot \frac{n(n + 1)}{2} + n \]

\[ S = -2n(n + 1) + n \]

Step 5: Simplify the Final Expression
Simplify the equation to find the final sum:

\[ S = -2n^2 - 2n + n \]

\[ S = -2n^2 - n \]

Conclusion:
By pairing consecutive terms and simplifying, the sum of the series \(1^2 - 2^2 + 3^2 - 4^2 + \ldots\) up to \(2n\) terms is found to be \(-2n^2 - n\), aligning with the solution derived through segregating positive and negative terms.