# Sequences and Series – Exercise 9.4 – Class XI

1. Find the sum of n terms of the series 1×2 + 2×3 + 3×4 + 4×5 +…

Solution:

The given series is 1×2 + 2×3 + 3×4 + 4×5 +… nth term, an=n(n+1) 1. Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5+ …

Solution:

The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … n th term,

an = n ( n + 1) ( n + 2)

= (n2 + n) (n + 2)

= n3 + 3n2 + 2n 1. Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + …

Solution:

The given series is 3 ×12 + 5 × 22 + 7 × 32 + … nth term,

an = ( 2n + 1) n2 = 2n3 + n2 1. Find the sum to n terms of the series 1/1×2 + 1/2×3 + 1/3×4 + …

Solution:

The given series is 1/1×2 + 1/2×3 + 1/3×4 + …

nth term, an = 1/n(n+1) = 1/n1/n+1

a1 = 1/11/2

a2 = 1/21/3

a3 = 1/31/4

Adding the above terms column wise, we get

a1 + a2 + a3 + … +an = [1/1 + 1/2 + 1/3 +…+1/n] – [1/2 + 1/3 +…+1/n+1]

Sn = 1 – 1/n+1 = n+1-1/n+1 = n/n+1

1. Find the sum to n terms of the series 52 + 62 + 72 +…+202

Solution:

The given series 52 + 62 + 72 +…+202 nth terms

an = (n+4)2 = n2 + 8n + 16 = 1496 +1088+256

=2840

Therefore, 52 + 62 + 72 + …+202 = 2840

1. Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Solution:

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

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

= (3n) (3n + 5)

= 9n2 + 15n 1. Find the sum to n terms of the series 12 + (12 + 22)+ (12 + 22+32)+…

Solution:

The given series 12 + (12 + 22)+ (12 + 22+32)+… an = 12 + 22+32+…+n2 \

1. Find the sum to n terms of the series whose nth term is given by n (n + 1)(n + 4).

Solution:

an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n 1. Find the sum to n terms of the series whose nth terms is given by n2 + 2n

Solution:

an = n2 + 2n 1. Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

Solution:

an = (2n – 1)2 = 4n2 – 4n + 1 