Chapter 4 Sequences and Series Ex 4.5 Solutions

We know that,

5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + ….. upto n terms Now, 5, 7, 9, 11, … are in A.P. rth term = 5 + (r – 1) (2) = 2r + 3 7, 9, 11,. … are in A.P. rth term = 7 + (r – 1) (2) = 2r + 5 ∴ 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms

22 + 42 + 62 + 82 + …… upto n terms = (2 × 1)2 + (2 × 2)2 + (2 × 3)2 + (2 × 4)2 + ……

Let S = (702 – 692) + (682 – 672) + …… +(22 – 12) ∴ S = (22 – 12) + (42 – 32) + …… + (702 – 692) Here, 2, 4, 6,…, 70 is an A.P. with rth term = 2r and 1, 3, 5,….., 69 in A.P. with rth term = 2r – 1

1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3) Now, 1, 3, 5, 7, … are in A.P. with a = 1 and d = 2. ∴ rth term = 1 + (r – 1)2 = 2r – 1 3, 5, 7, 9, … are in A.P. with a = 3 and d = 2 ∴ rth term = 3 + (r – 1)2 = 2r + 1 and 5, 7, 9, 11, … are in A.P. with a = 5 and d = 2 ∴ rth term = 5 + (r – 1)2 = 2r + 3 ∴ 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… upto n terms = n(n + 1)[2n(n + 1) + 4n + 2 – 1] – 3n = n(n + l)(2n2 + 6n + 1) – 3n = n(2n3 + 8n2 + 7n + 1 – 3) = n(2n3 + 8n2 + 7n – 2)