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`        Factorise.     SjsjsjsjsnnsnsbsbsbshshshshdhdhsshdhhshJzjjz `
one month ago

1655 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 numberkindly approve :)
```
one month ago
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• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions