Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Show that 9^n + 1 - 8n - 9 is divisible by 64, whenever n is a positive integer.

Q13. Show that 9^{n+1} - 8n - 9 is divisible by 64, whenever n is a positive integer.

Answers (1)

If we want to prove that  9^{n+1} - 8n - 9is divisible by 64, then we have to prove that  9^{n+1} - 8n - 9=64k

As we know from binomial theorem, 

(1+x)^m=^mC_0+^mC_1x+^mC_2x^2+^mC_3x^3+....^mC_mx^m

By putting x = 8 and replacing m by n+1, we get,

9^{n+1}=^{n+1}C_0+\:^{n+1}C_18+^{n+1}C_28^2+.......+^{n+1}C_{n+1}8^{n+1}

9^{n+1}=1+8(n+1)+8^2(^{n+1}C_2+\:^{n+1}C_38+^{n+1}C_48^2+.......+^{n+1}C_{n+1}8^{n-1})

9^{n+1}=1+8n+8+64(k)

Now, using this,

9^{n+1} - 8n - 9=9+8n+64k-9-8n=64k

Hence, 9^{n+1} - 8n - 9 is divisible by 64.

Posted by

neha

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4
anam-khan

When will CBSE Board exam 2025 date sheet be released for Class 10, 12?

Latest Question