Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Factorise.     Sjsjsjsjsnnsnsbsbsbshshshshdhdhsshdhhsh
Jzjjz

9 months ago

2014 Points

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 :)

9 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions