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alisha [4.7K]
2 years ago
15

I need it within tomorrow please

Mathematics
1 answer:
Alenkinab [10]2 years ago
6 0

Any even number raised to any power will remain even. (e.g. 2² = 4, 2³ = 8, etc)

Any odd number raised to any power will remain odd. (e.g. 1² = 1³ = ... = 1, 3⁴ = 81, etc)

So (n+1)^{2^k}-n^{2^k} will always be even, since both

[even] - [odd] = [odd]

and

[odd] - [even] = [odd]

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Check the procedures in the picture attached.

1. Since we are using direct substitution, the only thing we need to do is replace x by 6 (the quantity x is approaching to) in our quantity to get:
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2. Again, since we are using direct substitution, th only thing we need to do is replace x by 3 (the quantity x is approaching to) to get:
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2e^{ \frac{ \pi }{2} } cos \frac{ \pi }{2}
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2e^{ \frac{ \pi }{2} } (0)=0
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