    1. Chapter 8 Class 11 Binomial Theorem
2. Serial order wise
3. Ex 8.1

Transcript

Ex 8.1, 13 Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer. Introduction Numbers divisible by 64 are 64 = 64 × 1 128 = 64 × 2 640 = 64 × 10 Any number divisible by 64 = 64 × Natural number Hence, In order to show that 9n+1 – 8n – 9 is divisible by 64, We have to prove that 9n+1 – 8n – 9 = 64k , where k is some natural number Ex 8.1, 13 Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer. In order to show that 9n+1 – 8n – 9 is divisible by 64, We have to prove that 9n+1 – 8n – 9 = 64k , where k is some natural number Writing (9)n+1 = (1 + 8) n+1 (9)n + 1 = n + 1C0 1(n+1) 80 + n + 1C1 1(n+1) – 1 (8)1 + n + 1 C2 1n – 2(8)2 +…… + n + 1Cn + 1 1(n+1) – (n+1) (8) n + 1 (9)n + 1 = n + 1C0 + n + 1C1 8 + n + 1 C2 (8)2 +…… + n + 1Cn + 1 (8) n + 1 = 1 + ﷐﷐𝑛 + 1﷯!﷮1!﷐𝑛 + 1 − 1﷯!﷯ (8) + n + 1 C2(8)2 + n + 1 C3 (8)3 +…… + 1 (8) n + 1 = 1 + ﷐﷐𝑛 + 1﷯(𝑛)!﷮﷐𝑛﷯!﷯ (8) + n + 1 C2(8)2 + n + 1 C3 (8)3 +…… + (8) n + 1 = 1 + (n + 1) (8) + n + 1 C2(8)2 + n + 1 C3 (8)3 +…… + (8) n + 1 = 1 + (8n + 8) + n + 1 C2(8)2 + n + 1 C3 (8)3 +…… + (8) n + 1 = 8n + 9 + n + 1 C2(8)2 + n + 1 C3 (8)3 + …… + (8) n + 1 Hence, 9n + 1 = 8n + 9 + n + 1 C2(8)2 + n + 1 C3 (8)3 + …… + (8) n + 1 9n + 1 – 8n – 9 = n + 1 C2(8)2 + n + 1 C3 (8)3 + …… + (8) n + 1 Taking 82 common from right side 9n + 1 – 8n – 9 = (8)2 ﷐n + 1 C2 + n + 1 C3 (8)3 – 2 + …… + (8) n + 1– 2﷯ 9n + 1 – 8n – 9 = (8)2 ﷐n + 1 C2 + n + 1 C3 (8)1 + …… + (8) n – 1﷯ 9n + 1 – 8n – 9 = 64 ﷐n + 1 C2 + n + 1 C3 (8)1 + …… + (8) n – 1﷯ 9n + 1 – 8n – 9 = 64k where k =﷐n + 1 C2 + n + 1 C3 (8)1 + …… + (8) n – 1﷯ is a natural number Thus , 9n + 1 – 8n – 9 is divisible by 64, Hence proved

Ex 8.1 