**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 type of questions which needed in solving the maths question paper in the exams.

**Exercise 9.1- Sequence and Series**

**Exercise 9.2 – Sequence and Series**

**Exercise 9.3 – Sequence and Series**

**NCERT solutions of 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 |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**Study notes of Maths and Science NCERT and CBSE from class 9 to 12**

**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 of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

**NCERT Solutions of class 9 science **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

**NCERT Solutions of class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT solutions of class 10 science**

**Solutions of class 10 last years Science question papers**

**CBSE Class 10 – Question paper of science 2020 with solutions**

**CBSE class 10 -Latest sample paper of science**

**NCERT solutions of class 12 maths**

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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