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dedylja [7]
3 years ago
13

Show that 3 · 4^n + 51 is divisible by 3 and 9 for all positive integers n.

Mathematics
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

Step-by-step explanation:

Hello, please consider the following.

3\cdot 4^n+51=3\cdot 4^n+3\cdot 17=3(4^n+17)

So this is divisible by 3.

Now, to prove that this is divisible by 9 = 3*3 we need to prove that

4^n+17 is divisible by 3. We will prove it by induction.

Step 1 - for n = 1

4+17=21= 3*7 this is true

Step 2 - we assume this is true for k so 4^k+17 is divisible by 3

and we check what happens for k+1

4^{k+1}+17=4\cdot 4^k+17=3\cdot 4^k + 4^k+17

3\cdot 4^k is divisible by 3 and

4^k+17 is divisible by 3, by induction hypothesis

So, the sum is divisible by 3.

Step 3 - Conclusion

We just prove that 4^n+17 is divisible by 3 for all positive integers n.

Thanks

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Choose the correct simplification of x to the 12th power times z to the 11th power all over x to the 2nd power times z to the 4t
Umnica [9.8K]

Answer:

<h3>The option A) x^{10}z^7 is correct answer.</h3><h3>The correct simplification for the given expression \frac{x^{12}\times z^{11}}{x^2\times z^4} is x^{10}z^7</h3>

Step-by-step explanation:

Given expression is x to the 12th power times z to the 11th power all over x to the 2nd power times z to the 4th power.

The given expression can be written as \frac{x^{12}\times z^{11}}{x^2\times z^4}

<h3>To choose the correct simplification of the given expression :</h3>

Now we have to simplify the given expression as below

\frac{x^{12}\times z^{11}}{x^2\times z^4}

=x^{12}\times z^{11}(x^{-2}\times z^{-4})   ( by using the identity \frac{1}{a^m}=a^{-m} )

=(x^{12}.x^{-2})\times (z^{11}.z^{-4})

=x^{12-2}\times z^{11-4} ( by using the identity a^m.a^n=a^{m+n} )

=x^{10}\times z^7

=x^{10}z^7

∴ \frac{x^{12}\times z^{11}}{x^2\times z^4}=x^{10}z^7

<h3>The correct simplification for the given expression \frac{x^{12}\times z^{11}}{x^2\times z^4} is x^{10}z^7</h3><h3>Hence option A) x^{10}z^7 is correct answer.</h3>

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