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Let's say I have the following sequence:

$$48, 98, 144, 196, 240, 294, 336$$

The sequence of first differences between terms is: $50, 46, 52, 44, 54, 42$

The sequence of second differences between terms is: $-4, +6, -8, +10, -12$

What I'm asking for help with is that obviously there is a constant difference between the sequence of second differences when negative sign is ignored, but is it possible to use this information to obtain a formula for the nth term of the original sequence?

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  • $\begingroup$ A careful use of $(-1)^n$ should do the trick. $\endgroup$ – TZakrevskiy Dec 22 '16 at 11:59
  • $\begingroup$ Also see here: quora.com/… $\endgroup$ – Rohan Dec 22 '16 at 12:00
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Let $a_0,a_1,a_2,\ldots $ be the sequence.

Here $a_0=48, a_1=98, a_2=144$.

From the differences, note that $a_{n+3}-2a_{n+2}+a_{n+1}=-(a_{n+2}-2a_{n+1}+a_{n})+2(-1)^{n+2}$

which implies $$a_{n+3}=a_{n+2}+a_{n+1}-a_n+2(-1)^n, n=0,1,2,\ldots$$

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Hint

Divide the last sequence in two parts:

odd terms: $-4,-8,-12,...$

even terms: $6,10,14,...$

Both sequence are arithmetic.

$1)$ odd terms:

$$a_{2k-1}=2+4k \quad \text{for} \quad k \ge 1$$

$2)$ even terms:

$$a_{2k}=-4k \quad \text{for} \quad k \ge 1$$

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