# Ex 8.1,13

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.