**NCERT solutions for class 11 maths exercise 9.4 of chapter 9-Sequence and Series**

NCERT solutions for class 11 maths exercise 9.4 of chapter 9-Sequence and Series are created for helping the students of 11 class students in clearing their doubts. These NCERT solutions are very important for the students of 11 class students for a better understanding of chapter 9-Sequence and Series. Study of these NCERT solutions will give you a technical idea of solving different types of questions needed in solving the maths question paper in the exams.

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**NCERT solutions for class 11 maths of chapter 9-Sequence and Series**

**Exercise 9.1- Sequence and Series**

**Exercise 9.2 – Sequence and Series**

**Exercise 9.3 – Sequence and Series**

**Find the sum to n terms of each of the series in Exercises 1 to 7.**

**Q1. 1 ×2 + 2 × 3 + 3 × 4 + 4 × 5 +……..**

Ans. Observing the first and second number of each term of the series,we can get n^{th} term a_{n }of the series as follows.

a_{n} = n^{th} term of (1,2,3……) **× **n^{th} term of (2,3,4……)

= [1+ (n-1)×1]× [2 + (n -1)×1]

= n × (n +1)

= n² + n

The sum of the series S_{n }is

S_{n}= ∑(n² + n)

= ∑n² + ∑n

**Q2. 1 × 2 ×3 + 2× 3×4+ 3× 4 ×5+……….**

Ans. Observing the first , second and third number of each term of the series,we can get n^{th} term a_{n }of the series as follows.

a_{n} = n^{th} term of (1,2,3……) **× **n^{th} term of (2,3,4……) ×n^{th} term of (3,4,5……)

= [1+ (n-1)×1]× [2 + (n -1)×1]× [3 + (n -1)1]

n(n + 1)(n + 2) = n(n² +3n +2) = n³ + 3n² +2n

The sum of the series S_{n }is

S_{n}=∑ (n³ + 3n² +2n)

= ∑n³ + 3∑n²+ 2∑n

Factorizing n² +5n +6

n² + 2n+3n + 6

n (n + 2) + 3(n +2)

(n + 2)(n +3)

**Q3.3 ×1² + 5 × 2² + 7 × 3²+……**

Ans. Observing the first and second number of each term of the series,we can get n^{th} term a_{n }of the series as follows.

a_{n} = n^{th} term of (3,5,7……) **× **n^{th} term of (1²,2²,3²……)

= [3+ (n-1)×2]× n²

= (2n + 1)n²= 2n³ + n²

The sum of the series S_{n }is

S_{n } = ∑(2n³ + n²)

= 2∑n³ +∑ n²

Ans. Observing the pattern of both numbers of denominators of each term.

we can get n^{th} term a_{n }of the series as follows.

We can write it in the form

Now,putting n =1,2,3…,we get the series

On adding the terms are canceled out diagonally, we get the sum of the series S_{n} as follows.

**Q5. 5² + 6² + 7²+……..20²**

Ans. The given series is 5² + 6² + 7²+……..20²

The sum S_{n} of the series (5² + 6² + 7²+……..20²) = Sum of the square of first 20 natural number – Sum of the square of first 4 natural number

5² + 6² + 7²+……..20² =( 1² + 2² +3²+4²+ 5²……..20²)-(1² + 2² +3²+4²)

The sum of the square of first natural number is given as

S_{n}= 2870 – 30 = 2840

**Q6. 3 × 8 + 6 × 11 + 9 × 14+……….**

Ans. The given series is 3 × 8 + 6 × 11 + 9 × 14+……….

The n^{th} term of the given series is

= n^{th} term of (3,6,9…..) × n^{th} term of (8,11,14…..)

=[3+(n-1)3] × [8 +(n-1)3]

=3n(3n +5) = 9n² +15n

The sum S_{n} of the series is

S_{n} = ∑(9n² +15n)

= 9∑n² +15∑n

S_{n}= 3n(n+1)(n+3)

**Q7. 1² + (1² +2²) +(1² +2²+ 3²) + ……**

Ans. We are given the series

1² + (1² +2²) +(1² +2²+ 3²) + ……

The n^{th} term of the given series is

(1² +2²+ 3²+……n²)

=∑n²

Factorizing n² +3n +2 =n² +2n+n +2 =n(n+2) +1(n+2) =(n+2)(n+1)

**Q8. n(n+1)(n+4)**

Ans. We are given the expression of n^{th} term of the series

n(n+1)(n+4)

Solving it

=n(n² +5n +4)

=n³ + 5n² + 4n

The sum S_{n} of the series is

S_{n} = ∑n³ + 5∑n² + 4∑n

Factorizing the expression

3n² +23n +34

=3n² + 6n+17n+34

=3n(n+2) + 17(n +2)

=(n+2)(3n +17)

Therefore the sum S_{n}

**Q9. n² +2 ^{n}**

Ans. We are given n^{th} term of the series

n² +2^{n}

Putting n =1,2,3…we get subsequent terms of the series as follows

a_{1} = 1^{2}+2^{1}

a_{2} = 2^{2}+ 2^{2}

a_{3} = 3^{2} +2^{3}

……………

………….

On adding all the terms we get the sum S_{n}

S_{n}= 1^{2}+2^{1 } + 2^{2}+ 2^{2} + 3^{2} +2^{3}…….

= (1² + 2² +3³ +….) + (2¹ +2² + 2³ +…..)

Since (2¹ +2² + 2³ +…..) is a GP

**Q10.(2n -1)²**

Ans.We are given the expression of n^{th} term of the series

(2n-1)²

Expading it

=4n² -4n +1

The sum S_{n} of the series is

S_{n} = 4∑n² – 4∑n +∑1 (∑1= 1+1+1….upto n terms =n)

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions for class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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