0

Guest

Courses New

Factorise. Sjsjsjsjsnnsnsbsbsbshshshshdhdhsshdhhsh Jzjjz

Factorise.     Sjsjsjsjsnnsnsbsbsbshshshshdhdhsshdhhsh
Jzjjz
 

Question Image
Grade:12

1 Answers

Aditya Gupta
2081 Points
4 years ago
we can write 2^(2^n) – 1 as [2^{2^(n – 1)}]^2 – 1 = [2^{2^(n – 1)} – 1][2^{2^(n – 1)} + 1]....(1)
now, by simple recursion you can again factorise 2^{2^(n – 1)} – 1, and then again.... so on till 1.
so, complete factorisation:
[2^{2^(n – 1)} + 1][2^{2^(n – 2)} + 1][2^{2^(n – 3)} + 1]........[2^{2^0}+1][2^{2^0} – 1]
or [2^{2^(n – 1)} + 1][2^{2^(n – 2)} + 1][2^{2^(n – 3)} + 1]........[2^{2^0}+1].....(2)
(1) is a shorter factorisation while (2) is a complete factorisation. from (2) we can also infer that 2^(2^n) – 1 is always divisible by 3 if n is a natural number
kindly approve :)
 

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More