Sum to n Terms of Special Series

Q: What is the Vn method?

The V n method is finding the difference between two consecutive terms in terms of n and at the end only two terms that is one first term and one last term in terms of n are left.

Q: Give the formula to find the sum of n terms of an AP.

Answer: The sum, S n of n terms of an AP with the first term ‘a’ and common difference ‘d’ is given by <img alt="\begin{array}{l}{S_{n}=\frac{n}{2}[2 a+(n-1) d]} \\ {\text { OR }} \\ {S_{n}=\frac{n}{2}[a+l]} \\ {a \rightarrow \text { first term }} \\ {d \rightarrow \text { common difference }} \\ {n \rightarrow \text { number of terms }}\end{array}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cbegin%7Barray%7D%7Bl%7D%7BS_%7Bn%7D%3D%5Cfrac%7Bn%7D%7B2%7D%5B2%20a+%28n-1%29%20d%5D%7D%20%5C%5C%20%7B%5Ctext%20%7B%20OR%20%7D%7D%20%5C%5C%20%7BS_%7Bn%7D%3D%5Cfrac%7Bn%7D%7B2%7D%5Ba+l%5D%7D%20%5C%5C%20%7Ba%20%5Crightarrow%20%5Ctext%20%7B%20first%20term%20%7D%7D%20%5C%5C%20%7Bd%20%5Crightarrow%20%5Ctext%20%7B%20common%20difference%20%7D%7D%20%5C%5C%20%7Bn%20%5Crightarrow%20%5Ctext%20%7B%20number%20of%20terms%20%7D%7D%5Cend%7Barray%7D" class="fr-fic fr-dii">

Q: If a1, a2, a3............are in AP. then what is the value of a1a2 + a2a3...................anan+1 and also a1a2a3+ a2a3a4......................+anan+1 an+2 ?

Answer: <img alt="\begin{array}{l}{\text { If } a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots, a_{n} \text { are in AP. Then }} \\ {\qquad \begin{aligned} \bullet & a_{1} a_{2}+a_{2} a_{3}+\ldots \ldots \ldots+a_{n} a_{n+1}=\frac{1}{3\left(a_{2}-a_{1}\right)}\left(a_{n} a_{n+1} a_{n+2}-a_{0} a_{1} a_{2}\right) \\ \bullet & a_{1} a_{2} a_{3}+a_{2} a_{3} a_{4}+\ldots \ldots \ldots+a_{n} a_{n+1} a_{n+2}=\frac{1}{4\left(a_{2}-a_{1}\right)}\left(a_{n} a_{n+1} a_{n+2} a_{n+2}-a_{0} a_{1} a_{2} a_{4}\right) \end{aligned}}\end{array}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20If%20%7D%20a_%7B1%7D%2C%20a_%7B2%7D%2C%20a_%7B3%7D%2C%20%5Cldots%20%5Cldots%20%5Cldots%2C%20a_%7Bn%7D%20%5Ctext%20%7B%20are%20in%20AP.%20Then%20%7D%7D%20%5C%5C%20%7B%5Cqquad%20%5Cbegin%7Baligned%7D%20%5Cbullet%20%26%20a_%7B1%7D%20a_%7B2%7D+a_%7B2%7D%20a_%7B3%7D+%5Cldots%20%5Cldots%20%5Cldots+a_%7Bn%7D%20a_%7Bn+1%7D%3D%5Cfrac%7B1%7D%7B3%5Cleft%28a_%7B2%7D-a_%7B1%7D%5Cright%29%7D%5Cleft%28a_%7Bn%7D%20a_%7Bn+1%7D%20a_%7Bn+2%7D-a_%7B0%7D%20a_%7B1%7D%20a_%7B2%7D%5Cright%29%20%5C%5C%20%5Cbullet%20%26%20a_%7B1%7D%20a_%7B2%7D%20a_%7B3%7D+a_%7B2%7D%20a_%7B3%7D%20a_%7B4%7D+%5Cldots%20%5Cldots%20%5Cldots+a_%7Bn%7D%20a_%7Bn+1%7D%20a_%7Bn+2%7D%3D%5Cfrac%7B1%7D%7B4%5Cleft%28a_%7B2%7D-a_%7B1%7D%5Cright%29%7D%5Cleft%28a_%7Bn%7D%20a_%7Bn+1%7D%20a_%7Bn+2%7D%20a_%7Bn+2%7D-a_%7B0%7D%20a_%7B1%7D%20a_%7B2%7D%20a_%7B4%7D%5Cright%29%20%5Cend%7Baligned%7D%7D%5Cend%7Barray%7D" class="fr-fic fr-dii">

Sum to n Terms of Special Series

Edited By Komal Miglani | Updated on Oct 11, 2024 10:19 AM IST

If the terms of a sequence follow some pattern that can be defined by an explicit formula in n, then the sequence is called a progression. There are some sequences that do not form any series even after finding the difference of successive terms they do not form any series . So, in that case, we use the V_nMethod. In real life, we use the V_nMethod for finding the sum of a special series.

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The sum of some special series (Vn method)
Solved Examples Based on the Vn Method
Summary

Sum to n Terms of Special Series

In this article, we will cover the V_nMethod. This category falls under the broader category of sequence and series, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main Exam (from 2013 to 2023), a total of eleven questions have been asked on this concept, including two in 2013, two in 2021, four in 2022, and two in 2023.

The sum of some special series (V_n method)

The V_n method is finding the difference between two consecutive terms in terms of n and at the end only two terms that is one first term and one last term in terms of n are left.

Case 1: Sum of the series of the form

$\begin{array}{l} a_{1} a_{2} \dots \dots a_{r} + a_{2} a_{3} \dots \dots a_{r + 1} + \dots \dots \dots + a_{n} a_{n + 1} \dots \dots a_{n + r - 1} \\ S_{n} = a_{1} a_{2} \dots \dots a_{r} + a_{2} a_{3} \dots \dots a_{r + 1} + \dots \dots \dots + a_{n} a_{n + 1} \dots \dots a_{n + r - 1} \\ T_{n} = a_{n} a_{n + 1} \dots \dots \dots \dots a_{n + r - 2} a_{n + r - 1} \\ Let V_{n} = a_{n} a_{n + 1} \dots \dots \dots \dots a_{n + r - 2} a_{n + r - 1} a_{n + r} [Taking one extra factor in T_{n} for V_{n}] \\ V_{n - 1} = a_{n - 1} a_{n} a_{n + 1} \dots \dots \dots a_{n + r - 3} a_{n + r - 2} a_{n + r - 1} \\ \Rightarrow V_{n} - V_{n - 1} = a_{n} a_{n + 1} a_{n + 2} \dots \dots a_{n + r - 1} (a_{n + r} - a_{n - 1}) = T_{n} (a_{n + r} - a_{n - 1}) \end{array}$

d be the common difference of an AP, then

$\begin{array}{l} a_{n} = a_{1} + (n - 1) d \\ V_{n} - V_{n - 1} = T_{n} [{a_{1} + (n + r - 1) d} - {a_{1} + (n - 2) d}] = (r + 1) d \cdot T_{n} \\ \Rightarrow T_{n} = \frac{1}{(r + 1) d} (V_{n} - V_{n - 1}) \\ S_{n} = \sum T_{n} = \frac{1}{(r + 1) d} \sum_{n = 1}^{n} (V_{n} - V_{n - 1}) \\ S_{n} = \frac{1}{(r + 1) d} (V_{n} - V_{0}) [from the telescoping series] \\ S_{n} = \frac{1}{(r + 1) (a_{2} - a_{1})} (a_{n} a_{n + 1} a_{n + 2} \dots a_{n + r} - a_{0} a_{1} a_{2} \dots \dots a_{r}) \end{array}$

$\begin{array}{l} If a_{1}, a_{2}, a_{3}, \dots \dots \dots, a_{n} are in AP. Then \\ \begin{aligned} ∙ & a_{1} a_{2} + a_{2} a_{3} + \dots \dots \dots + a_{n} a_{n + 1} = \frac{1}{3 (a_{2} - a_{1})} (a_{n} a_{n + 1} a_{n + 2} - a_{0} a_{1} a_{2}) \\ ∙ & a_{1} a_{2} a_{3} + a_{2} a_{3} a_{4} + \dots \dots \dots + a_{n} a_{n + 1} a_{n + 2} = \frac{1}{4 (a_{2} - a_{1})} (a_{n} a_{n + 1} a_{n + 2} a_{n + 2} - a_{0} a_{1} a_{2} a_{4}) \end{aligned} \end{array}$

Case 2: Sum of the series of the form

$\begin{array}{l} \frac{1}{a_{1} a_{2} \dots a_{r}} + \frac{1}{a_{2} a_{3} \dots a_{r + 1}} + \dots \dots + \frac{1}{a_{n} a_{n + 1} \dots a_{n + r - 1}} \\ S_{n} = \frac{1}{a_{1} a_{2} \dots \dots a_{r}} + \frac{1}{a_{2} a_{3} \dots a_{r + 1}} + \dots \dots + \frac{1}{a_{n} a_{n + 1} \dots \dots a_{n + r - 1}} \\ T_{n} = \frac{1}{a_{n} a_{n + 1} \dots \dots \dots a_{n + r - 2} a_{n + r - 1}} \\ Let, V_{n} = \frac{1}{a_{n + 1} a_{n + 2} \dots \dots a_{n + r - 2} a_{n + r - 1}} [Leaving first factor from the denominator of T_{n}] \\ So, V_{n - 1} = \frac{1}{a_{n} a_{n + 1} \dots \dots \dots a_{n + r - 3} a_{n + r - 2}} \end{array}$

$\begin{array}{l} \Rightarrow V_{n} - V_{n - 1} = \frac{1}{a_{n + 1} a_{n + 2} \dots \dots \dots a_{n + r - 2} a_{n + r - 1}} - \frac{1}{a_{n} a_{n + 1} \dots \dots \dots a_{n + r - 3} a_{n + r - 2}} \\ T_{n} = \frac{(V_{n} - V_{n - 1})}{d (1 - r)} \\ S_{n} = \sum T_{n} = \sum_{n = 1}^{n} \frac{(V_{n} - V_{n - 1})}{d (1 - r)} = \frac{1}{d (1 - r)} (V_{n} - V_{0}) \\ S_{n} = \frac{1}{(r - 1) (a_{2} - a_{1})} {\frac{1}{a_{1} a_{2} \dots a_{r - 1}} - \frac{1}{a_{n + 1} a_{n + 2} \dots a_{n + r - 1}}} \end{array}$

$\begin{array}{l} If a_{1}, a_{2}, a_{3}, \dots \dots \dots, a_{n} are in AP. Then \\ \begin{aligned} ∙ & \frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots \dots + \frac{1}{a_{n} a_{n + 1}} = \frac{n}{a_{1} a_{n + 1}} \\ ∙ \frac{1}{a_{1} a_{2} a_{3}} + \frac{1}{a_{2} a_{3} a_{4}} + \dots \dots + \frac{1}{a_{n} a_{n + 1} a_{n + 2}} = \frac{1}{2 (a_{2} - a_{1})} {\frac{1}{a_{1} a_{2}} - \frac{1}{a_{n + 1} a_{n + 2}}} \end{aligned} \end{array}$

Solved Examples Based on the V_nMethod

Example 1: If $a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + \dots \dots + a_{25}$ is equal to: [JEE MAINS 2023]

Solution: Given that $a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15}$

$\begin{aligned} a_{1} + a_{2} + a_{3} + \dots . a_{25} = \sum_{n = 1}^{25} \frac{- 2}{(2 n - 3) (2 n - 5)} \\ = \sum_{n = 1}^{25} \frac{(2 n - 5) - (2 n - 3)}{(2 n - 3) (2 n - 5)} \\ = \sum_{n = 1}^{25} (\frac{1}{2 n - 3} - \frac{1}{(2 n - 5)}) \\ = \frac{1}{- 1} - \frac{1}{- 3} \end{aligned}$

$\begin{aligned} + \frac{1}{1} - \frac{1}{- 1} \\ + \frac{1}{3} - \frac{1}{1} \\ ⋮ ⋮ \\ \frac{1}{47} - \frac{1}{45} \\ = \frac{1}{47} + \frac{1}{3} \\ = \frac{3 + 47}{141} = \frac{50}{141} \end{aligned}$

Hence, the answer is $\frac{50}{141}$

Example 2: The sum $\sum_{n = 1}^{21} \frac{3}{(4 n - 1) (4 n + 3)}$ is equal to [JEE MAINS 2022]

Solution: Required sum

$= \frac{3}{3.7} + \frac{3}{7.11} + \dots + \frac{3}{83.87}$

$= \frac{3}{4} (\frac{4}{3.7} + \frac{4}{7.11} + \dots + \frac{4}{83.87})$

$= \frac{3}{4} (\frac{7 - 3}{3.7} + \frac{11 - 7}{7.11} + \dots + \frac{87 - 85}{83.87})$

$= \frac{3}{4} (\frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \dots + \frac{1}{83} - \frac{1}{87})$

$= \frac{3}{4} (\frac{1}{3} - \frac{1}{87})$

$= \frac{3}{4} (\frac{29 - 1}{87})$

$= \frac{3.28}{4 \cdot 87}$

$= \frac{7}{29}$

Hence, the answer is $\frac{7}{29}$

Example 3: $\sum_{r = 1}^{20} (r^{2} + 1) (r!)$ is equal to: [JEE MAINS 2022]

Solution

$\begin{aligned} (r^{2} + 1) r!_{0} \\ = & (γ^{2} + 2 r + 1 - 2 r) r!_{0} \\ = & ((r + 1)^{2} - 2 r) r! \\ = & (γ + 1)^{2} \cdot r! - 2 r \cdot r! \\ = & (r + 1) \cdot (r + 1)!_{0} - γ \cdot γ!_{0} - γ \cdot r! \\ = & [(r + i) (γ + 1)! - γ \cdot γ!_{0}] - [((r + 1) - 1) \cdot γ!] \\ = & [(r + 1) (γ + 1)! - γ \cdot γ!] - [(γ + 1)! - γ!] \\ Applying summation from 1 to 20 \\ = (21.21! - 1) - (21! - 1) \\ = 20 \cdot 21! \\ = (22 - 2) 21! \\ = 22! - 2 \cdot 21! \end{aligned}$

Hence, the answer is 22! - 2. 21!

Example 4: If $\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \frac{1}{4 \times 5 \times 6} + \dots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}$ , then $34 k$ is equal to________ [JEE MAINS 2022]

Solution

$\begin{aligned} \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \dots + \frac{1}{100 \cdot 101 \cdot 102} = \frac{k}{101} \\ \frac{4 - 2}{2 \cdot 3 \cdot 4} + \frac{5 - 3}{3 \cdot 4 \cdot 5} + \dots + \frac{102 - 100}{100 \cdot 101 \cdot 102} = \frac{2 k}{101} \\ \frac{1}{2.3} - \frac{1}{3 \cdot 4} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \dots \frac{1}{100 \cdot 101} - \frac{1}{101 \cdot 102} = \frac{2 k}{101} \\ \frac{1}{2 \cdot 3} - \frac{1}{101 \cdot 102} = \frac{2 k}{101} \\ ∴ 2 k = \frac{101}{6} - \frac{1}{102} \\ ∴ 34 k = 286 \end{aligned}$

Hence, the answer is the 286.

Example 5: If $\frac{1}{(20 - a) (40 - a)} + \frac{1}{(40 - a) (60 - a)} + \dots + \frac{1}{(180 - a) (200 - a)} = \frac{1}{256}$ , then the maximum value of a is: [JEE MAINS 2022]

Solution

$By slitting \begin{aligned} \frac{1}{20} [(\frac{1}{20 - a} - \frac{1}{40 - a}) + (\frac{1}{40 - a} - \frac{1}{60 - a}) + \dots . \dots + (\frac{1}{180 - a} - \frac{1}{200 - a})] \\ \Rightarrow \frac{1}{20} (\frac{1}{20 - a} - \frac{1}{200 - a}) = \frac{1}{256} \\ (20 - a) (200 - a) = 256 \times 9 \\ a^{2} - 220 a + 1696 = 0 \\ a = 8, 212 \\ Hence maximum value of a is 212 . \end{aligned}$

Hence, the answer is 212.

Summary

V_n method helps us to find the sum of special series that are neither in AP nor in GP. Understanding of V_n method helps us to know the pattern of the series and also enables us to find the general term of the sequence. This method remains valuable in computational mathematics for its ability to handle a wide range of sequences and series, enhancing our understanding and application of numerical techniques in mathematical analysis.

Frequently Asked Questions (FAQs)

1. What is the Vn method?

The V_n method is finding the difference between two consecutive terms in terms of n and at the end only two terms that is one first term and one last term in terms of n are left.

2. What is the geometric sequence?

A geometric sequence is a sequence where the first term is non-zero and the ratio between consecutive terms is always constant. The ‘constant factor’ is called the common ratio and is denoted by ‘r’. r is also a non-zero number.

3. What is arithmetic progression?

An arithmetic progression is a sequence in which each term increases or decreases by a constant term or fixed number. This fixed number is called the common difference of an AP and is generally denoted by ‘d’.

4. Give the formula to find the sum of n terms of an AP.

Answer: The sum, S_n of n terms of an AP with the first term ‘a’ and common difference ‘d’ is given by

$\begin{array}{l}{S_{n}=\frac{n}{2}[2 a+(n-1) d]} \\ {\text { OR }} \\ {S_{n}=\frac{n}{2}[a+l]} \\ {a \rightarrow \text { first term }} \\ {d \rightarrow \text { common difference }} \\ {n \rightarrow \text { number of terms }}\end{array}$

5. If a1, a2, a3............are in AP. then what is the value of a1a2 + a2a3...................anan+1 and also a1a2a3+ a2a3a4......................+anan+1 an+2 ?

Answer:

$\begin{array}{l}{\text { If } a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots, a_{n} \text { are in AP. Then }} \\ {\qquad \begin{aligned} \bullet & a_{1} a_{2}+a_{2} a_{3}+\ldots \ldots \ldots+a_{n} a_{n+1}=\frac{1}{3\left(a_{2}-a_{1}\right)}\left(a_{n} a_{n+1} a_{n+2}-a_{0} a_{1} a_{2}\right) \\ \bullet & a_{1} a_{2} a_{3}+a_{2} a_{3} a_{4}+\ldots \ldots \ldots+a_{n} a_{n+1} a_{n+2}=\frac{1}{4\left(a_{2}-a_{1}\right)}\left(a_{n} a_{n+1} a_{n+2} a_{n+2}-a_{0} a_{1} a_{2} a_{4}\right) \end{aligned}}\end{array}$