Class 12 Mathematics SEQUENCE AND SERIES Notes

Class 12 Mathematics SEQUENCE AND SERIES Notes
Class 12 Mathematics Sequence and Series Notes | Nepal eNotes

Unit 6

Algebra

Class 12 Mathematics

Sequence and Series

Class 12 Mathematics – Sequence and Series Notes PDF

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Contents / Quick Navigation

1. Sequence and Series 2. Sum of Natural Numbers 3. Sum of Squares 4. Sum of Cubes 5. Mathematical Induction 6. Induction Proofs of Sum Formulae 7. Applications 8. Worked Examples 9. Common Mistakes 10. Important Exam Questions 11. One-Minute Revision 12. Diagram Practice

NEB / CDC Focus

For this Class 12 unit, the central results are the sums of the first n natural numbers, their squares and their cubes, together with the principle of mathematical induction and its use in proving finite-sum identities.

1. Sequence and Series

Sequence

A sequence is an ordered list of numbers or terms: a₁,a₂,a₃,….

Series

A series is obtained by adding the terms of a sequence: a₁+a₂+a₃+….

Syllabus Focus

The Class 12 unit does not require a fresh treatment of arithmetic and geometric progressions here. The focus is on standard finite sums and mathematical induction.

Sequence vs Series Sequence 1, 2, 3, 4, …, n an ordered list Series 1+2+3+4+…+n the sum of listed terms add
Figure 1: A sequence lists terms; a series adds them.

2. Sum of the First n Natural Numbers

Let

S=1+2+3+…+(n−1)+n.

Write the same sum in reverse:

S=n+(n−1)+(n−2)+…+2+1.

Adding term by term:

2S=(n+1)+(n+1)+…+(n+1).

There are n equal terms on the right, so

Sum of Natural Numbers
1+2+3+…+n = n(n+1)/2.
Pairing Method 123 …n−2n−1n nn−1n−2 …321 Every vertical pair sums to n+1 2S=n(n+1) ⇒ S=n(n+1)/2
Figure 2: Reversing and pairing the series produces n identical sums of n+1.
Worked Example

Find 1+2+3+…+100.

S=100(101)/2=5050.

2.1 Sum from m to n

For positive integers m≤n,

m+(m+1)+…+n = n(n+1)/2 − (m−1)m/2.

3. Sum of the Squares of the First n Natural Numbers

Sum of Squares
1²+2²+3²+…+n² = n(n+1)(2n+1)/6.

This formula can be proved by mathematical induction. It is also useful in statistics, coordinate geometry and algebraic simplification.

1² + 2² + 3² + … + n² 1² 2² 3² + … + n² Each term k² is the number of unit squares in a k×k square.
Figure 3: A visual interpretation of square-number terms.
Worked Example

Find 1²+2²+…+20².

S=20(21)(41)/6=2870.

3.1 Sum of Squares from m to n

Subtract the sum through m−1 from the sum through n:

Σk=mnk² = n(n+1)(2n+1)/6 − (m−1)m(2m−1)/6.

4. Sum of the Cubes of the First n Natural Numbers

Sum of Cubes
1³+2³+3³+…+n³ = [n(n+1)/2]².

Thus the sum of the first n cubes is exactly the square of the sum of the first n natural numbers:

1³+2³+…+n³ = (1+2+…+n)².

A Beautiful Identity Sum of Cubes 1³+2³+…+n³ equals Square of Natural Sum [n(n+1)/2]² For n=4: 1+8+27+64=100=(1+2+3+4)²
Figure 4: The sum-of-cubes formula is the square of the triangular-number formula.
Worked Example

Find 1³+2³+…+10³.

S=[10(11)/2]²=55²=3025.

5. Principle of Mathematical Induction

Principle of Mathematical Induction

Suppose P(n) is a statement about positive integers. To prove P(n) for every positive integer n, it is sufficient to show:

  1. Base step: P(1) is true.
  2. Inductive step: whenever P(k) is true, P(k+1) is also true.

The assumption that P(k) is true is called the induction hypothesis.

Domino Idea Behind Mathematical Induction 123 kk+1… base starts chain k ⇒ k+1
Figure 5: A valid base case starts the chain, while the inductive step transfers truth from k to k+1.

5.1 Standard Proof Structure

  1. State the proposition P(n).
  2. Verify it for the initial value, usually n=1.
  3. Assume P(k) is true for an arbitrary positive integer k.
  4. Start the left side of P(k+1).
  5. Use the induction hypothesis to replace the part through k.
  6. Simplify until the right side becomes exactly the formula with n=k+1.
  7. Conclude that the statement holds for all positive integers.
Mathematical Induction Proof Flow Define the statement P(n) Verify P(1) Assume P(k) is true Prove P(k+1) using P(k) Therefore P(n) is true ∀n∈N
Figure 6: The standard structure of an induction proof.
Remember / Warning

You may assume only P(k), not the statement you are trying to prove for k+1. The inductive step must genuinely derive P(k+1) from the induction hypothesis.

6. Proving Standard Sums by Mathematical Induction

6.1 Proof of 1+2+…+n=n(n+1)/2

Proposition
P(n): 1+2+…+n=n(n+1)/2.

Base case: For n=1,

LHS=1,   RHS=1(2)/2=1.

Induction hypothesis: Assume for n=k,

1+2+…+k=k(k+1)/2.

For n=k+1:

1+2+…+k+(k+1) = k(k+1)/2+(k+1)

=(k+1)(k+2)/2.

This is exactly the required formula with n=k+1. Hence the formula holds for all positive integers.

6.2 Proof of the Sum of Squares

Proposition
1²+2²+…+n²=n(n+1)(2n+1)/6.

Base case: For n=1, both sides equal 1.

Assume

1²+2²+…+k²=k(k+1)(2k+1)/6.

Then

1²+2²+…+k²+(k+1)² = k(k+1)(2k+1)/6+(k+1)².

Factor k+1:

=(k+1)[k(2k+1)+6(k+1)]/6

=(k+1)(2k²+7k+6)/6

=(k+1)(k+2)(2k+3)/6.

Since 2k+3=2(k+1)+1, this is precisely the formula for k+1.

6.3 Proof of the Sum of Cubes

Proposition
1³+2³+…+n³=[n(n+1)/2]².

Base case: At n=1, both sides equal 1.

Assume

1³+2³+…+k³=[k(k+1)/2]².

Then

1³+2³+…+k³+(k+1)³ = k²(k+1)²/4+(k+1)³.

Factor:

=(k+1)²[k²/4+(k+1)]

=(k+1)²(k²+4k+4)/4

=(k+1)²(k+2)²/4 = [(k+1)(k+2)/2]².

Thus the formula is true for all positive integers.

Three Standard Finite Sums Natural Numbers Σk n(n+1)/2 Squares Σk² n(n+1)(2n+1)/6 Cubes Σk³ [n(n+1)/2]² All three identities are natural candidates for proof by mathematical induction.
Figure 7: Formula map for the core Class 12 results.

7. Applications of the Standard Sums

7.1 Sums Containing Linear Expressions

Example: Sum 3k+2

Find Σk=1n(3k+2).

=3Σk+2Σ1 =3[n(n+1)/2]+2n.

=n(3n+7)/2.

7.2 Sums Containing Quadratic Expressions

For a polynomial such as Ak²+Bk+C, split the sum:

Σ(Ak²+Bk+C)=AΣk²+BΣk+CΣ1.
Worked Example

Find Σk=1n(2k²+3k+1).

=2[n(n+1)(2n+1)/6] +3[n(n+1)/2] +n.

The expression may then be simplified according to the form required in the question.

7.3 Odd and Even Number Sums

1+3+5+…+(2n−1)=n² 2+4+6+…+2n=n(n+1)

These identities follow directly from the natural-number sum and are also excellent practice statements for mathematical induction.

Why 1+3+5+…+(2n−1)=n² 1 1+3=4=2² 1+3+5=9=3² … nth square Each next odd number adds an L-shaped border to form the next square.
Figure 8: Successive odd numbers build square numbers.

8. Worked Examples

Example 1: Sum from 25 to 100

Find 25+26+…+100.

=[1+2+…+100]−[1+2+…+24]

=100(101)/2−24(25)/2 =5050−300 =4750.

Example 2: Sum of squares from 6 to 15

Find 6²+7²+…+15².

=15(16)(31)/6−5(6)(11)/6

=1240−55=1185.

Example 3: Sum of cubes

Evaluate 3³+4³+…+12³.

=[1³+…+12³]−[1³+2³]

=[12(13)/2]²−[2(3)/2]² =78²−3² =6084−9 =6075.

Example 4: Induction – odd-number sum

Prove 1+3+5+…+(2n−1)=n².

Base: At n=1, 1=1².

Assume: 1+3+…+(2k−1)=k².

Then:

1+3+…+(2k−1)+[2(k+1)−1] =k²+(2k+1) =(k+1)².

Hence the result is true for every positive integer n.

Example 5: Induction – divisibility

Prove that 7ⁿ−1 is divisible by 6 for all positive integers n.

Base: 7¹−1=6, divisible by 6.

Assume: 7ᵏ−1=6m for some integer m.

Then

7k+1−1 =7·7ᵏ−1 =7(7ᵏ−1)+6 =7(6m)+6 =6(7m+1).

Thus it is divisible by 6.

Example 6: Find n from a natural-number sum

If 1+2+…+n=210, find n.

n(n+1)/2=210 ⇒ n²+n−420=0.

(n−20)(n+21)=0.

Since n is positive, n=20.

Example 7: Polynomial sum

Find Σk=1n(k²+k).

=Σk²+Σk =n(n+1)(2n+1)/6+n(n+1)/2.

=n(n+1)[(2n+1)+3]/6 =n(n+1)(n+2)/3.

Example 8: Verify the cube identity numerically

For n=5,

1³+2³+3³+4³+5³=1+8+27+64+125=225.

[5(6)/2]²=15²=225.

The two sides agree.

9. Problem-Solving Strategy

  1. Identify whether the sum involves k, k², or k³.
  2. For a range beginning above 1, use “sum through n minus sum through m−1”.
  3. Split polynomial sums using linearity: Σ(Ak²+Bk+C)=AΣk²+BΣk+CΣ1.
  4. In induction, write the base case explicitly.
  5. State the induction hypothesis exactly before proving P(k+1).
  6. When proving a sum identity, begin the k+1 case by adding the next term to the sum through k.
  7. Factor aggressively in the inductive step so the result takes the required k+1 form.
  8. Do a quick numerical check for a small value of n to catch algebraic errors.

10. Common Mistakes and Warnings

Mistake 1: Wrong square-sum denominator

The formula is n(n+1)(2n+1)/6, not divided by 2 or 3.

Mistake 2: Forgetting the square in the cube sum

Σk³=[n(n+1)/2]².

Mistake 3: Wrong lower-limit subtraction

For a sum from m to n, subtract the sum through m−1, not through m.

Mistake 4: Skipping the base case

An induction proof is incomplete without verifying the starting value.

Mistake 5: Assuming P(k+1)

The induction hypothesis is P(k); P(k+1) must be proved.

Mistake 6: Not adding the next term

For a finite sum, the k+1 case is the sum through k plus the new (k+1) term.

Mistake 7: Using induction for n=0 without checking

Use the starting integer specified by the proposition.

Mistake 8: Treating verification as proof

Checking several numerical cases supports a pattern but does not replace mathematical induction.

11. Exam-Important Formula Sheet

Core Formulas
Σk=1n1 = n Σk=1nk = n(n+1)/2 Σk=1nk² = n(n+1)(2n+1)/6 Σk=1nk³ = [n(n+1)/2]² 1+3+5+…+(2n−1)=n² 2+4+6+…+2n=n(n+1) Induction: P(1) true and P(k)⇒P(k+1) imply P(n) true for all positive integers n.

12. Important Exam Questions

Short-Answer Questions

  1. Define a sequence and a series.
  2. State the formula for 1+2+…+n.
  3. State the formula for 1²+2²+…+n².
  4. State the formula for 1³+2³+…+n³.
  5. State the principle of mathematical induction.
  6. What is an induction hypothesis?
  7. Find the sum of the first n even natural numbers.
  8. Find the sum of the first n odd natural numbers.

Derivation / Proof Questions

  1. Derive 1+2+…+n=n(n+1)/2 by the pairing method.
  2. Prove by mathematical induction that 1+2+…+n=n(n+1)/2.
  3. Prove by mathematical induction that 1²+2²+…+n²=n(n+1)(2n+1)/6.
  4. Prove by mathematical induction that 1³+2³+…+n³=[n(n+1)/2]².
  5. Use induction to prove a supplied finite-sum identity.
  6. Use induction to prove a simple divisibility statement valid for all positive integers.

Numerical / Application Questions

  1. Evaluate a finite sum of consecutive natural numbers.
  2. Evaluate a finite range of squares or cubes.
  3. Find n when a standard finite sum is given.
  4. Evaluate a polynomial sum involving k and k².
  5. Use the standard formulas to simplify a finite summation.

Diagram / Concept Questions

  1. Use a pairing diagram to explain the sum of the first n natural numbers.
  2. Use the domino idea to explain the principle of mathematical induction.
  3. Use a square-building diagram to explain 1+3+5+…+(2n−1)=n².

13. One-Minute Revision

Quick Revision
  • A sequence is an ordered list; a series is the sum of sequence terms.
  • 1+2+…+n=n(n+1)/2.
  • 1²+2²+…+n²=n(n+1)(2n+1)/6.
  • 1³+2³+…+n³=[n(n+1)/2]².
  • The cube sum is the square of the natural-number sum.
  • For a sum from m to n, subtract the sum through m−1.
  • 1+3+…+(2n−1)=n².
  • 2+4+…+2n=n(n+1).
  • Mathematical induction requires a base case.
  • The induction hypothesis assumes P(k) for an arbitrary positive integer k.
  • The inductive step must prove P(k+1).
  • For finite-sum induction, add the next term to the result through k.
  • Checking a few values is not a proof for all natural numbers.
  • Standard sum formulas are especially useful for polynomial summations.
  • Always simplify the final inductive expression into the exact required k+1 form.

14. Diagram Practice

  • Sequence versus series concept diagram.
  • Pairing method for 1+2+…+n.
  • Square-number array illustrating 1²+2²+…+n².
  • Identity map for the sum of cubes.
  • Domino analogy for mathematical induction.
  • Mathematical induction proof flowchart.
  • Formula map for sums of k, k² and k³.
  • Odd-number square-building diagram.

Also Visit

Class 12 Mathematics Notes Class 12 Notes Practical File Question Papers Important Questions
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Original Nepal eNotes page: Class 12 Mathematics Sequence and Series Notes.

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