My Math Forum Demonstrate that An^2 + Bn is an arithmetic sequence

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 April 14th, 2015, 05:42 PM #1 Senior Member   Joined: Jan 2015 From: USA Posts: 107 Thanks: 2 Demonstrate that An^2 + Bn is an arithmetic sequence The problem is as the title states: Demonstrate that An^2 + Bn is an arithmetic sequence. I don`t know how to solve this. Any ideas?
 April 14th, 2015, 06:17 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Its an arithmetic series, not an arithmetic sequence. Investigate $$b_n = a_n - a_{n-1} = An^2 + Bn - A(n-1)^2 - B(n-1)$$ This gives you the difference between consecutive partial sums of the series. You can then look at $b_n - b_{n-1}$ to find the difference between successive terms of the sequence. Thanks from topsquark and matisolla Last edited by v8archie; April 14th, 2015 at 06:20 PM.
April 14th, 2015, 06:50 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, matisolla!

Where did this problem come from?
It makes little sense.

Quote:
 Demonstrate that $An^2 + Bn$ is an arithmetic sequence. I don't know how to solve this. Any ideas? $\;\;{\color{blue}{\text{No!}}$

This is NOT an arithmetic sequence.

$\;\;\begin{array}{ccccc}
n && a(n) && d \\ \hline

1 && A+B \\

&&&& 3A+B \\

2 && 4A+2B \\
&&&& 5A+B \\

3 && 9A+3B \\
&&&& 7A+B \\

4 && 16A+4B \\

&& \vdots && \vdots

\end{array}$

The common difference, $d$, is not constant.

 Tags an2, arithmetic, demonstrate, sequence

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