Does not depend on n

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Suppose that $S(k)$ is the sum of the first k terms of an arithmetic sequence with common difference 3. If the value of$$S(3n) / S(n)$$ does not depend on n, what is the 100th term of the sequence?

What does "does not depend on n" mean and how do I make use of it in this problem?

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1
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"Does not depend on $n$" means that $S(3n)/S(n)$ gives the same value no matter what value you choose for $n$. It may depend on other things, but it is a constant regarding to $n$.

The function

$$f(n):=\frac{S(3n)}{S(n)}$$

is constant.


You can use this by setting

$$\frac{S(3n)}{S(n)}=\mathrm{const}=c\quad\implies\quad S(3n)=c\cdot S(n).$$

Hint: This helps you to deduce $S(3^k n)=c^k S(n)$.

0
On

$\frac {S(3n)}{S(n)}$ does not depend on $n$ means that the expression doesn't have any term containing $n$ in it. So basically you have to set up the conditions on other variable which is your first term of the sequence, say $a$ in such a way that $n$ vanishes.

But then, there you are, with first term and common difference which characterizes the whole sequence.

0
On

When we are talking about differentiable functions, we often use the property that $f' = 0 \Leftrightarrow f = \mathrm{const}$. We could try to use similar approach here. Let us define 'descrete derivative' as

$$ f'(n) = f(n) - f(n-1) $$

and 'discrete antiderivative' as

$$F(n) = \sum\limits^n f(n).$$

It is easy to see that $F'(n) = f(n)$. Our 'derivative' satisfies a property, that is very similar to $(f/g)' = f' / g - f g' / g^2$. That is

$$ \left(\frac{f}{g} \right)'(n) = \frac{f'(n)}{g(n)} - \frac{f(n-1) g'(n)}{g(n) g(n-1)}. $$

The property '$S(3m) / S(m)$ does not depend on $m$' could be read now as $(S(3m) / S(m))' = 0$.

Notice that $$ S'(3m) = a_{3m} + a_{3m-1} + a_{3m-2}. $$

All that's left is to solve the equation.