Sum to n Terms of Special Series

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

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

d be the common difference of an AP, then

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

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

Case 2: Sum of the series of the form

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

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

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

Solved Examples Based on the V_n Method

Example 1: If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\cdots \ldots+a_{25}$ is equal to: [JEE MAINS 2023]

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

$\begin{aligned} & a_1+a_2+a_3+\ldots . a_{25}=\sum_{n=1}^{25} \frac{-2}{(2 n-3)(2 n-5)} \\ & =\sum_{\mathrm{n}=1}^{25} \frac{(2 n-5)-(2 n-3)}{(2 n-3)(2 n-5)} \\ & =\sum_{\mathrm{n}=1}^{25}\left(\frac{1}{2 n-3}-\frac{1}{(2 n-5)}\right) \\ & =\frac{1}{-1}-\frac{1}{-3} \quad \end{aligned}$

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

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

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

$=\frac{3}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\cdots+\frac{4}{83.87}\right) \\$

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

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

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

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

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

$=\frac{7}{29}$

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

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

Solution

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

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

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

Solution

$\begin{aligned} &\mathrm{\frac{1 }{2\cdot3\cdot4}+\frac{1}{3 \cdot 4 \cdot 5}+\cdots+\frac{1}{100 \cdot 101 \cdot 102}=\frac{k}{101} }\\ &\mathrm{\frac{4-2}{2 \cdot 3 \cdot 4}+\frac{5-3}{3 \cdot 4 \cdot 5}+\cdots+\frac{102-100}{100 \cdot 101 \cdot 102}=\frac{2 k}{101}} \\ &\mathrm{\frac{1}{2.3}-\frac{1}{3 \cdot 4}+\frac{1}{3 \cdot 4}-\frac{1}{4 \cdot 5}+\cdots \frac{1}{100 \cdot 101}-\frac{1}{101 \cdot 102}=\frac{2 k}{101} }\\ &\mathrm{\frac{1}{2 \cdot 3}-\frac{1}{101 \cdot 102}=\frac{2 k}{101}} \\ &\mathrm{\therefore 2 k=\frac{101}{6}-\frac{1}{102}} \\ &\mathrm{\therefore 34 k=286} \end{aligned}$

Hence, the answer is the 286.

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

Solution

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

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.