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Chapter 8 Class 11 Binomial Theorem
Ex 8.1,4 Important
Example 5 Important Deleted for CBSE Board 2023 Exams
Ex 8.2,10 Important Deleted for CBSE Board 2023 Exams
Ex 8.2,12 Deleted for CBSE Board 2023 Exams
Example 13 Important Deleted for CBSE Board 2023 Exams
Example 15 Important Deleted for CBSE Board 2023 Exams
Misc 1 Important Deleted for CBSE Board 2023 Exams
Misc 8 Important Deleted for CBSE Board 2023 Exams
Chapter 8 Class 11 Binomial Theorem
Last updated at Jan. 29, 2020 by Teachoo
Ex 8.1, 13 - Introduction Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer. 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) + n + 1C1 1(n + 1) – 1 (8)1 + n + 1 C2 1n – 2(8)2 +…… + n + 1Cn + 1 (8) n + 1 We know that (a + b)n = nC0 an + nC1 an – 1 b 1 + nC2 an – 2 b2 + ….. + nCn bn Putting a = 1 ,b = 8 and n = 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