NCERT solutions for class 11 maths exercise 9.4 of chapter 9-Sequence and Series - Future Study Point

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

sequence and series -ncert

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

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 nth term an of the series as follows.

an =  nth term of (1,2,3……) × nth term of (2,3,4……)

= [1+ (n-1)×1]× [2 + (n -1)×1]

= n × (n +1)

= n² + n

The sum of the series Sn is

Sn=  ∑(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 nth term an of the series as follows.

an =  nth term of (1,2,3……) × nth term of (2,3,4……) ×nth 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 Sn is

Sn=∑ (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 nth term an of the series as follows.

an =  nth term of (3,5,7……) × nth term of (1²,2²,3²……)

= [3+ (n-1)×2]× n²

= (2n + 1)n²= 2n³ + n²

The sum of the series Sis

Sn  = ∑(2n³ + n²)

= 2∑n³ +∑ n²

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

we can get nth term an of the series as follows.

We can write it in the form

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

sequence and series

 

 

 

 

 

 

On adding the terms are canceled out diagonally, we get the sum of the series Sn  as follows.

Q5. 5² + 6² + 7²+……..20²

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

The sum Sn 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

Sn= 2870 – 30 = 2840

Q6. 3 × 8 + 6 × 11 + 9 × 14+……….

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

The nth term of the given series is

= nth term of (3,6,9…..) × nth term of (8,11,14…..)

=[3+(n-1)3] × [8 +(n-1)3]

=3n(3n +5) = 9n² +15n

The sum Sn of the series is

Sn = ∑(9n² +15n)

= 9∑n² +15∑n

Sn= 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 nth 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 nth term of the series

n(n+1)(n+4)

Solving it

=n(n² +5n +4)

=n³ + 5n² + 4n

The sum Sn of the series is

Sn = ∑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 Sn

Q9. n² +2n

Ans. We are given nth term of the series

n² +2n

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

a1 = 12+21

a2 = 22+ 22

a3 = 32 +23

……………

………….

On adding all the terms we get the sum Sn

Sn= 12+21   + 22+ 22 + 32 +23…….

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

Since (2¹ +2² + 2³ +…..) is a GP

Q10.(2n -1)²

Ans.We are given the expression of nth term of the series

(2n-1)²

Expading it

=4n² -4n +1

The sum Sn of the series is

Sn = 4∑n²  – 4∑n +∑1 (∑1= 1+1+1….upto n terms =n)

 

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Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
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Chapter 1-Relations and FunctionsChapter 9-Differential Equations
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