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Naddika [18.5K]
3 years ago
12

Given the proposition,

Mathematics
1 answer:
zubka84 [21]3 years ago
6 0

Answer and Explanation:

Given : P(n): 1 + 2 + 2^2 + 2^3 + . . . + 2^n = 2^{n+1} - 1, n=0,1,2,..

To find : The values of following expression ?

Solution :

The function is P(n)=2^{n+1} - 1

1) Value of P(0),

P(0)=2^{0+1} - 1

P(0)=2^{1} - 1

P(0)=2 - 1

P(0)=1

2) Value of P(1),

P(1)=2^{1+1} - 1

P(1)=2^{2} - 1

P(1)=4- 1

P(1)=3

3) Value of P(2),

P(2)=2^{2+1} - 1

P(2)=2^{3} - 1

P(2)=8- 1

P(2)=7

4) Value of P(n+1),

P(n+1)=2^{n+1+1} - 1

P(n+1)=2^{n+2} - 1

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Answer:

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Please help me with this problem <br> Evaluate (-½·3)³
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Answer: -27/8

Step-by-step explanation:

\left(-\frac{1}{2}\cdot \:3\right)^3

\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=-a^n,\:\quad \mathrm{if\:}n\mathrm{\:is\:odd}

\left(-\frac{1}{2}\cdot \:3\right)^3=-\left(\frac{1}{2}\cdot \:3\right)^3

=-\left(\frac{1}{2}\cdot \:3\right)^3

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