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Prove that if the sequence $a_{n}$ is arithmetic, then for any positive integer k the sequence: $b_{1} = a_{1}+a_{2} +...+ a_{k}, \\ b_{2}=a_{k+1}+...+ a_{2k}, \\ b_{3}= a_{2k+1}+...+ a_{3k}$ is arithmetic as well. What is the difference of the sequence $b_{n}$ ?

I figured that a given $b_n$ has a form of $b_n=a_{(n-1)k+1}+a_{(n-1)k+2}+...+a_{nk}$. Hence, $b_{n+1}=a_{nk+1}+a_{nk+2}+...+a_{(n+1)k}$. We know that $a_n$ is arithmetic so we may use the sum formula and we get: $b_{n+1}=\frac{a_{nk+1}+a_{(n+1)k}}{2}\cdot k \\ b_{n}=\frac{a_{(n-1)k+1}+a_{nk}}{2}\cdot k$ And further: $b_{n+1}-b_n=\frac{ka_{nk+1}+ka_{(n+1)k}-ka_{(n-1)k+1}-a_{nk}}2=\frac{1}{2}k(a_{nk+1}-a_{(n-1)k+1}+a_{(n+1)k}-a_{nk})$ And I believe it's enough to show that $b_n$ is arithmetic as well. $\frac{1}{2}k$ is constant, $a_{nk+1}-a_{(n-1)k+1}$ are two adjacent terms in an arithmetic progression so their difference is certainly constant and $a_{(n+1)k}-a_{(n-1)k}$ are separated by some term in the middle but as it's arithmetic sequence, $(a_{(n+1)k}-a_{nk}=const \land a_{nk}-a_{(n-1)k}=const)\rightarrow a_{(n+1)k}-a_{(n-1)k}=const.$ Is it enough to prove or this should be done some other way? [this paragraph may be a little off now as I found an error - there is no $a_{(n-1)k}$ here but $a_{nk}$ instead which I plugged into the equations above]

And as regards the difference - I thought about using the formula that $a_n=a_1+(n-1)d$ so we have: $b_{n+1}-b_n=\frac{1}{2}k([a_1+nkd]+[a_1+(nk+k-1)d]-[a_1+(nk-k)d]-[a_1+(nk-1)d])=\frac{k}{2}\cdot 2kd=k^2d$

So the difference of $b_n$ is $k^2d$. Is it OK?

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    You're absolutely right. Thank you for pointing that out!2012-05-01

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It could be organized more clearly, but basically it's fine. Here's how I'd do it.

Let $d$ be the common difference of the original sequence, i.e., $d=a_{n+1}-a_n$ for each $n$. For any $n$ we have $a_{n+k}-a_n=kd$, so for any $n$ we have

$\begin{align*} b_{n+1}-b_n&=\sum_{i=1}^ka_{nk+i}-\sum_{i=1}^ka_{(n-1)k+i}\\ &=\sum_{i=1}^k(a_{nk+i}-a_{(n-1)k+i})\\ &=\sum_{i=1}^kkd\\ &=k^2d\;. \end{align*}$

This is constant, so the sequence $\langle b_1,b_2,\dots\rangle$ is arithmetic with common difference $dk^2$.

Note that there's nothing wrong with using the sum formula as you did; it just adds some extra work.

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    Yeah, I know - I just don't feel confident with the sum sign yet and preferred to write it down that way :) Thank you.2012-05-01