Prove that for any natural n the number 3^(n+4)−3^n is divisible by 16.
1 answer:
Answer:
Proved (See Explanation)
Step-by-step explanation:
Show that 3ⁿ⁺⁴ - 3ⁿ is divisible by 16.
This is done as follows
From laws of indices;
aᵐ⁺ⁿ = aᵐ * aⁿ.
So, 3ⁿ⁺⁴ can be written as 3ⁿ * 3⁴.
becomes
Factorize
3ⁿ * 5
5(3ⁿ)
<em>The expression can not be further simplified.</em>
<em>However, we can conclude that when 3ⁿ⁺⁴ - 3ⁿ is divisible by 16, because 5(3ⁿ) is a natural whole number as long as n is a natural whole number.</em>
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Answer:
<em>Diverges; the limit of the terms, an, is not 0 as n goes to infinity</em>
Step-by-step explanation:
<u>Convergence of Infinite Series</u>
The series is given as the terms
1 , 9 , 1 , 18 , 27 , 1 , 36 , 1 , 45 ...
For a series to be convergent, the limit of the general term an when n goes to infinity must be 0. In such cases, the sum of all terms tends to a fixed value.
Our series has two clearly different sequences, which define it as a piecewise general term:
If n is odd, then
Otherwise, an is an arithmetic series starting from 9 with a common difference of 9, that is
if n is even.
If we take the limit as n goes to infinity, we don't know which one should be selected, thus we must assume that any of them is a correct option, therefore both of them should converge.
The limit of the function
is 1 when n goes to infinity. Since it's not 0, this piece is not convergent and the entire series isn't either.
But it goes worse because of the limit of the second piece
when n goes to infinity is infinite.
The answer is
<em>Diverges; the limit of the terms, an, is not 0 as n goes to infinity</em>