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_nof 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_nis

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_nof 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_nis

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_nof 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_nis

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_nof 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 +2ⁿ

Ans. We are given n^th 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 S_n

S_n= 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 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)

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