Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 3 down vote favorite

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?

share|improve this question
1  
You are working too hard to find the difference for the $b_n$. What is $a_{k+s}-a_s$? To get from $a_s$ to $a_{k+s}$ we added $d$ $k$ times. So $a_{k+1}-a_1=kd$, $a_{k+2}-a_2=kd$, and so on up to $a_{2k}-a_k=kd$, total $k^2d$. Same difference for the next bunch, forever. – André Nicolas May 1 '12 at 18:12
OK, thank you :) – Straightfw May 1 '12 at 18:27
1  
$a_{nk+1}-a_{(n-1)k+1}$ is not the difference of adjacent terms of $a$ – Thomas Andrews May 1 '12 at 19:17
You're absolutely right. Thank you for pointing that out! – Straightfw May 1 '12 at 19:19

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
Yeah, I know - I just don't feel confident with the sum sign yet and preferred to write it down that way :) Thank you. – Straightfw May 1 '12 at 18:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.