Chapter 9: Sequences and Series Exercise 9.4

Given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

It’s seen that,

n^th term, a_n = n ( n + 1)

Then, the sum of n terms of the series can be expressed as

Given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

It’s seen that,

n^th term, a_n = n ( n + 1) ( n + 2)

= (n² + n) (n + 2)

= n³ + 3n² + 2n

Then, the sum of n terms of the series can be expressed as

Given series is 3 ×1² + 5 × 2² + 7 × 3² + …

It’s seen that,

n^th term, a_n = ( 2n + 1) n² = 2n³ + n²

Then, the sum of n terms of the series can be expressed as

Given series is 5² + 6² + 7² + … + 20²

It’s seen that,

n^th term, a_n = ( n + 4)² = n² + 8n + 16

Then, the sum of n terms of the series can be expressed as

Given series is 3 × 8 + 6 × 11 + 9 × 14 + …

It’s found out that,

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

= (3n) (3n + 5)

= 9n² + 15n

Then, the sum of n terms of the series can be expressed as

Given series is 1² + (1² + 2²) + (1² + 2² + 3² ) + …

Finding the n^th term, we have

a_n = (1² + 2² + 3² +…….+ n²)

Now, the sum of n terms of the series can be expressed as

Given,

a_n = n (n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n

Now, the sum of n terms of the series can be expressed as

Given,

n^th term of the series as:

a_n = n² + 2ⁿ

Then, the sum of n terms of the series can be expressed as

Given,

n^th term of the series as:

a_n = (2n – 1)² = 4n² – 4n + 1

Then, the sum of n terms of the series can be expressed as