Special Series (AGP and Method of Difference)

Introduction to Special Series

Beyond AP and GP, there are other important types of series that frequently appear in competitive exams. This section covers methods for summing Arithmetico-Geometric Progressions and series using sigma notation and the method of difference.

Arithmetico-Geometric Progression (AGP)

An AGP is a series where each term is the product of the corresponding terms of an AP and a GP.

• General Form: $a, (a+d)r, (a+2d)r^2, ..., (a+(n-1)d)r^{n-1}$

• Method for Summing an AGP:

1. Let $S_n$ be the sum of the n terms.
2. Multiply the entire series by the common ratio 'r' and shift the terms by one position.
3. Subtract the second series from the first: $S_n - rS_n = (1-r)S_n$.
4. The resulting series will be a GP, which can be summed easily. Finally, solve for $S_n$.

• Sum of an Infinite AGP: If $|r|<1$, the sum to infinity is given by:

$$S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$$

Summation using Sigma ($\Sigma$) Notation

For series where the n-th term ($a_n$) can be expressed as a polynomial in 'n', we can find the sum using standard summation formulas:

• Sum of first n natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
• Sum of squares of first n natural numbers: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
• Sum of cubes of first n natural numbers: $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$

Method of Difference (Telescoping Series)

This method is used when the differences between consecutive terms are in AP or GP. The goal is to express the general term $a_k$ as a difference of two consecutive terms of another sequence, i.e., $a_k = V_{k} - V_{k-1}$. When summed, the series telescopes:

$$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (V_k - V_{k-1}) = (V_1-V_0) + (V_2-V_1) + ... + (V_n-V_{n-1}) = V_n - V_0$$

Example 1: Sum of an Infinite AGP

Question: Find the sum of the infinite series $1 + \frac{4}{5} + \frac{7}{25} + \frac{10}{125} + ...$

Solution: The numerators 1, 4, 7, 10… are in AP with $a=1, d=3$. The denominators 1, 5, 25, 125… are in GP with $r=1/5$. Since $|r|<1$, the sum exists. Using the formula $S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$:

$$S_{\infty} = \frac{1}{1-1/5} + \frac{3(1/5)}{(1-1/5)^2} = \frac{1}{4/5} + \frac{3/5}{(4/5)^2} = \frac{5}{4} + \frac{3/5}{16/25}$$

$$= \frac{5}{4} + \frac{3}{5} \cdot \frac{25}{16} = \frac{5}{4} + \frac{15}{16} = \frac{20+15}{16} = \frac{35}{16}$$

Example 2: Summation using $\Sigma$ Notation

Question: Find the sum of the first n terms of the series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ...$

Solution: The n-th term is $a_n = n(n+1) = n^2+n$. The sum is $S_n = \sum_{k=1}^{n} (k^2+k) = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$. Using the standard formulas:

$$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$$

$$= \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] = \frac{n(n+1)}{2} \left[ \frac{2n+4}{3} \right] = \frac{n(n+1)(n+2)}{3}$$

Example 3: Method of Difference (Telescoping Series)

Question: Find the sum of n terms of the series $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ...$

Solution: The n-th term is $a_n = \frac{1}{n(n+1)}$. We can split this using partial fractions: $a_n = \frac{1}{n} - \frac{1}{n+1}$. Now we write out the terms of the sum: $S_n = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{n} - \frac{1}{n+1})$. The intermediate terms cancel out (telescoping sum).

$$S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}$$