📘 Chapter 4: Sequence and Series | NEB Mathematics Notes Class 12
Sum of series, nth term, special series & summation formulas
🔢 Sequence and Series covers finding the nth term and sum of various series including arithmetico-geometric series, sum of squares, cubes, and natural numbers. Below are detailed, line-by-line solutions to typical NEB examination problems.
📌 Important Summation Formulas:
\(\sum n = \frac{n(n+1)}{2}\) | \(\sum n^2 = \frac{n(n+1)(2n+1)}{6}\) | \(\sum n^3 = \left[\frac{n(n+1)}{2}\right]^2\)
\(\sum (a + (r-1)d) = \frac{n}{2}[2a + (n-1)d]\) | \(\sum ar^{r-1} = \frac{a(r^n – 1)}{r-1}\)
\(\sum n = \frac{n(n+1)}{2}\) | \(\sum n^2 = \frac{n(n+1)(2n+1)}{6}\) | \(\sum n^3 = \left[\frac{n(n+1)}{2}\right]^2\)
\(\sum (a + (r-1)d) = \frac{n}{2}[2a + (n-1)d]\) | \(\sum ar^{r-1} = \frac{a(r^n – 1)}{r-1}\)
📊 1. Sum of Series (Given nth term)
1d \(t_n = 3^n – 4n^3\), find \(S_n\)
Step 1: \(S_n = \sum t_n = \sum 3^n – \sum 4n^3\)
Step 2: \(\sum 3^n = \frac{3(3^n – 1)}{3 – 1} = \frac{3(3^n – 1)}{2}\) (Geometric series)
Step 3: \(\sum 4n^3 = 4\left[\frac{n(n+1)}{2}\right]^2 = 4 \cdot \frac{n^2(n+1)^2}{4} = n^2(n+1)^2\)
Step 4: \(S_n = \frac{3(3^n – 1)}{2} – n^2(n+1)^2\)
Step 2: \(\sum 3^n = \frac{3(3^n – 1)}{3 – 1} = \frac{3(3^n – 1)}{2}\) (Geometric series)
Step 3: \(\sum 4n^3 = 4\left[\frac{n(n+1)}{2}\right]^2 = 4 \cdot \frac{n^2(n+1)^2}{4} = n^2(n+1)^2\)
Step 4: \(S_n = \frac{3(3^n – 1)}{2} – n^2(n+1)^2\)
✅ \(S_n = \frac{3(3^n – 1)}{2} – n^2(n+1)^2\)
🔢 2. Series: \(1\cdot1 + 2\cdot2 + 3\cdot3 + \dots\)
2a \(1\cdot1 + 2\cdot2 + 3\cdot3 + \dots\)
\(t_n = n \times n = n^2\)
\(S_n = \sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
\(S_n = \sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
✅ \(\frac{n(n+1)(2n+1)}{6}\)
2b \(1\cdot3 + 2\cdot4 + 3\cdot5 + \dots\)
\(t_n = n \times (n+2) = n^2 + 2n\)
\(S_n = \sum n^2 + 2\sum n = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}\)
\(= \frac{n(n+1)(2n+1)}{6} + n(n+1) = \frac{n(n+1)(2n+1 + 6)}{6} = \frac{n(n+1)(2n+7)}{6}\)
\(S_n = \sum n^2 + 2\sum n = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}\)
\(= \frac{n(n+1)(2n+1)}{6} + n(n+1) = \frac{n(n+1)(2n+1 + 6)}{6} = \frac{n(n+1)(2n+7)}{6}\)
✅ \(\frac{n(n+1)(2n+7)}{6}\)
2c \(1\cdot3 + 3\cdot5 + 5\cdot7 + \dots\)
nth term of 1,3,5,… = \(2n-1\)
nth term of 3,5,7,… = \(2n+1\)
\(t_n = (2n-1)(2n+1) = 4n^2 – 1\)
\(S_n = 4\sum n^2 – \sum 1 = 4 \cdot \frac{n(n+1)(2n+1)}{6} – n = \frac{2n(n+1)(2n+1)}{3} – n\)
\(= n\left[\frac{2(n+1)(2n+1)}{3} – 1\right] = \frac{n}{3}(4n^2 + 6n – 1)\)
nth term of 3,5,7,… = \(2n+1\)
\(t_n = (2n-1)(2n+1) = 4n^2 – 1\)
\(S_n = 4\sum n^2 – \sum 1 = 4 \cdot \frac{n(n+1)(2n+1)}{6} – n = \frac{2n(n+1)(2n+1)}{3} – n\)
\(= n\left[\frac{2(n+1)(2n+1)}{3} – 1\right] = \frac{n}{3}(4n^2 + 6n – 1)\)
✅ \(\frac{n}{3}(4n^2 + 6n – 1)\)
2d \(2\cdot5 + 4\cdot8 + 6\cdot11 + \dots\)
nth term of 2,4,6,… = \(2n\)
nth term of 5,8,11,… = \(5 + (n-1)3 = 3n + 2\)
\(t_n = 2n(3n+2) = 6n^2 + 4n\)
\(S_n = 6\sum n^2 + 4\sum n = 6 \cdot \frac{n(n+1)(2n+1)}{6} + 4 \cdot \frac{n(n+1)}{2}\)
\(= n(n+1)(2n+1) + 2n(n+1) = n(n+1)(2n+1+2) = n(n+1)(2n+3)\)
nth term of 5,8,11,… = \(5 + (n-1)3 = 3n + 2\)
\(t_n = 2n(3n+2) = 6n^2 + 4n\)
\(S_n = 6\sum n^2 + 4\sum n = 6 \cdot \frac{n(n+1)(2n+1)}{6} + 4 \cdot \frac{n(n+1)}{2}\)
\(= n(n+1)(2n+1) + 2n(n+1) = n(n+1)(2n+1+2) = n(n+1)(2n+3)\)
✅ \(n(n+1)(2n+3)\)
2e \(5\cdot3 + 7\cdot5 + 9\cdot7 + \dots\)
nth term of 5,7,9,… = \(5 + (n-1)2 = 2n+3\)
nth term of 3,5,7,… = \(3 + (n-1)2 = 2n+1\)
\(t_n = (2n+3)(2n+1) = 4n^2 + 8n + 3\)
\(S_n = 4\sum n^2 + 8\sum n + 3n = 4 \cdot \frac{n(n+1)(2n+1)}{6} + 8 \cdot \frac{n(n+1)}{2} + 3n\)
\(= \frac{2n(n+1)(2n+1)}{3} + 4n(n+1) + 3n\)
After simplification: \(\frac{n}{3}(4n^2 + 12n + 11)\)
nth term of 3,5,7,… = \(3 + (n-1)2 = 2n+1\)
\(t_n = (2n+3)(2n+1) = 4n^2 + 8n + 3\)
\(S_n = 4\sum n^2 + 8\sum n + 3n = 4 \cdot \frac{n(n+1)(2n+1)}{6} + 8 \cdot \frac{n(n+1)}{2} + 3n\)
\(= \frac{2n(n+1)(2n+1)}{3} + 4n(n+1) + 3n\)
After simplification: \(\frac{n}{3}(4n^2 + 12n + 11)\)
✅ \(\frac{n}{3}(4n^2 + 12n + 11)\)
✨ 3. Special Series
3a \(1 + (1+3) + (1+3+5) + \dots\)
\(1 = 1^2\), \(1+3 = 2^2\), \(1+3+5 = 3^2\), …
\(t_n = n^2\)
\(S_n = \sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
\(t_n = n^2\)
\(S_n = \sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
✅ \(\frac{n(n+1)(2n+1)}{6}\)
3b \(1\cdot n + 2\cdot(n-1) + 3\cdot(n-2) + \dots\)
\(t_r = r \cdot (n – r + 1) = nr – r^2 + r\)
\(S_n = n\sum r – \sum r^2 + \sum r = n \cdot \frac{n(n+1)}{2} – \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\)
\(= \frac{n(n+1)}{2}\left[n – \frac{2n+1}{3} + 1\right] = \frac{n(n+1)}{2} \cdot \frac{3n – 2n – 1 + 3}{3} = \frac{n(n+1)(n+2)}{6}\)
\(S_n = n\sum r – \sum r^2 + \sum r = n \cdot \frac{n(n+1)}{2} – \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\)
\(= \frac{n(n+1)}{2}\left[n – \frac{2n+1}{3} + 1\right] = \frac{n(n+1)}{2} \cdot \frac{3n – 2n – 1 + 3}{3} = \frac{n(n+1)(n+2)}{6}\)
✅ \(\frac{n(n+1)(n+2)}{6}\)
3c \(1 + (1+3) + (1+3+5) + \dots\) to n terms
Same as 3a. Sum of first n odd numbers = \(n^2\), so series becomes \(1^2 + 2^2 + 3^2 + \dots + n^2\)
\(S_n = \frac{n(n+1)(2n+1)}{6}\)
\(S_n = \frac{n(n+1)(2n+1)}{6}\)
✅ \(\frac{n(n+1)(2n+1)}{6}\)
📈 4. Series involving Sum of Natural Numbers
4a \(2 + (2+4) + (2+4+6) + \dots\)
Sum of first n even numbers = \(n(n+1)\)
Here, 1st term = 2, 2nd term = 2+4 = 6, 3rd term = 2+4+6 = 12, …
So \(t_n = n(n+1) = n^2 + n\)
\(S_n = \sum n^2 + \sum n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1 + 3)}{6} = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}\)
Here, 1st term = 2, 2nd term = 2+4 = 6, 3rd term = 2+4+6 = 12, …
So \(t_n = n(n+1) = n^2 + n\)
\(S_n = \sum n^2 + \sum n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1 + 3)}{6} = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}\)
✅ \(\frac{n(n+1)(n+2)}{3}\)
4b \(1 + (1+2) + (1+2+3) + \dots\)
Sum of first n natural numbers = \(\frac{n(n+1)}{2}\)
So \(t_n = \frac{n(n+1)}{2} = \frac{1}{2}(n^2 + n)\)
\(S_n = \frac{1}{2}\left[\sum n^2 + \sum n\right] = \frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] = \frac{n(n+1)}{4}\left[\frac{2n+1}{3} + 1\right] = \frac{n(n+1)(n+2)}{6}\)
So \(t_n = \frac{n(n+1)}{2} = \frac{1}{2}(n^2 + n)\)
\(S_n = \frac{1}{2}\left[\sum n^2 + \sum n\right] = \frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] = \frac{n(n+1)}{4}\left[\frac{2n+1}{3} + 1\right] = \frac{n(n+1)(n+2)}{6}\)
✅ \(\frac{n(n+1)(n+2)}{6}\)
🔍 5. Finding nth term from pattern
5a Natural numbers grouped: (1), (2,3), (4,5,6), (7,8,9,10), …
First term of nth group: It is 1 more than sum of first (n-1) natural numbers
First term = \(1 + \frac{(n-1)n}{2} = \frac{n(n-1)}{2} + 1\)
Sum of nth group = sum of n consecutive numbers starting from first term
\(S_n = \frac{n}{2} \times [2 \times \text{first} + (n-1) \times 1] = \frac{n}{2}[n(n-1) + 2 + n – 1] = \frac{n}{2}(n^2 – n + n + 1) = \frac{n(n^2 + 1)}{2}\)
First term = \(1 + \frac{(n-1)n}{2} = \frac{n(n-1)}{2} + 1\)
Sum of nth group = sum of n consecutive numbers starting from first term
\(S_n = \frac{n}{2} \times [2 \times \text{first} + (n-1) \times 1] = \frac{n}{2}[n(n-1) + 2 + n – 1] = \frac{n}{2}(n^2 – n + n + 1) = \frac{n(n^2 + 1)}{2}\)
✅ First term = \(\frac{n(n-1)}{2} + 1\), Sum = \(\frac{n(n^2 + 1)}{2}\)
⚡ 6. Miscellaneous Series
6a \(2\cdot3 + 3\cdot4 + 4\cdot5 + \dots\)
\(t_n = (n+1)(n+2) = n^2 + 3n + 2\)
\(S_n = \sum n^2 + 3\sum n + 2n = \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} + 2n\)
\(= \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + 2n = \frac{n}{6}[(n+1)(2n+1) + 9(n+1) + 12]\)
\(= \frac{n}{6}(2n^2 + 12n + 22) = \frac{n}{3}(n^2 + 6n + 11)\)
\(S_n = \sum n^2 + 3\sum n + 2n = \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} + 2n\)
\(= \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + 2n = \frac{n}{6}[(n+1)(2n+1) + 9(n+1) + 12]\)
\(= \frac{n}{6}(2n^2 + 12n + 22) = \frac{n}{3}(n^2 + 6n + 11)\)
✅ \(\frac{n}{3}(n^2 + 6n + 11)\)
7 \(2,3 + 3,4 + 4,5 + \dots\)
\(t_n = (n+1)(n+2) = n^2 + 3n + 2\) (same as 6a)
\(S_n = \frac{n}{3}(n^2 + 6n + 11)\)
\(S_n = \frac{n}{3}(n^2 + 6n + 11)\)
✅ \(\frac{n}{3}(n^2 + 6n + 11)\)
8 \(2 + (2+4) + (2+4+6) + \dots\)
Sum of first n even numbers = \(n(n+1)\)
So \(t_n = n(n+1) = n^2 + n\)
\(S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}\)
So \(t_n = n(n+1) = n^2 + n\)
\(S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}\)
✅ \(\frac{n(n+1)(n+2)}{3}\)
9 \(1^3 + 2^3 + 3^3 + \dots\) (sum of cubes)
\(t_n = n^3\)
\(S_n = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}\)
\(S_n = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}\)
✅ \(\frac{n^2(n+1)^2}{4}\)
10 \(1^2 + 2^2 + 3^2 + \dots\) (sum of squares)
\(t_n = n^2\)
\(S_n = \frac{n(n+1)(2n+1)}{6}\)
\(S_n = \frac{n(n+1)(2n+1)}{6}\)
✅ \(\frac{n(n+1)(2n+1)}{6}\)
11 \(1 + 3 + 7 + 15 + 31 + \dots\)
Pattern: \(1 = 2^1 – 1\), \(3 = 2^2 – 1\), \(7 = 2^3 – 1\), \(15 = 2^4 – 1\), …
\(t_n = 2^n – 1\)
\(S_n = \sum (2^n – 1) = (2 + 4 + 8 + \dots + 2^n) – n = \frac{2(2^n – 1)}{2-1} – n = 2^{n+1} – 2 – n\)
\(t_n = 2^n – 1\)
\(S_n = \sum (2^n – 1) = (2 + 4 + 8 + \dots + 2^n) – n = \frac{2(2^n – 1)}{2-1} – n = 2^{n+1} – 2 – n\)
✅ \(S_n = 2^{n+1} – n – 2\)
📝 7. Additional Problem
12 Find sum: \(1\cdot3 + 2\cdot4 + 3\cdot5 + \dots\) to n terms
Same as 2b. \(t_n = n(n+2) = n^2 + 2n\)
\(S_n = \frac{n(n+1)(2n+1)}{6} + n(n+1) = \frac{n(n+1)(2n+7)}{6}\)
\(S_n = \frac{n(n+1)(2n+1)}{6} + n(n+1) = \frac{n(n+1)(2n+7)}{6}\)
✅ \(\frac{n(n+1)(2n+7)}{6}\)
🧠 Sequence and Series Key Formulas
📌 Sum of first n natural numbers: \(\frac{n(n+1)}{2}\)
📌 Sum of squares: \(\frac{n(n+1)(2n+1)}{6}\)
📌 Sum of cubes: \(\left[\frac{n(n+1)}{2}\right]^2\)
📌 Sum of first n even numbers: \(n(n+1)\)
📌 Sum of first n odd numbers: \(n^2\)
📌 Geometric series sum: \(\frac{a(r^n – 1)}{r-1}\) for \(r \neq 1\)