Same Mom trying to help daughter here. Once I know how to explain this pattern, can you tell me what I'd call this area of math so I can go re-teach myself? I feel like I could have done this in high school, but it's all gone from lack of use...
$2, 7, 26, 101, 400$
I've fooled around with some $a(x) + b(y) =$ differences, but nothing. Is this really middle school math? Thank you thank you all!
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$\begingroup$The other question that you posted was multiple choice. If this one is too, then there are probably some answers that can be eliminated.
Without meaning to discount others' comments about "there is no one right answer," given the scope of middle school math there are some reasonable things that can be said about the sequence $2,7,26,101,400,...$
- The next number should reasonably be bigger than $400$. We don't see any example given that suggests the contrary. (If $300$ were a choice, it could be eliminated.)
- The next number should be quite a bit bigger, given that the difference between consecutive elements in the sequence have a wider gap each time. (If $401$ were a choice, it could be reasonably eliminated.)
- The next number should be close to four times bigger, looking at what's presented. $101$ to $400$ is the strongest suggestion of this but it's reasonable for the other pairs of consecutive numbers as well.
Now, if the last point is noticed, then we can home in. Starting with $2$, multiplying by $4$ each time would give
$$2,8,32,128,512$$
Not quite right, so let's look at the first two numbers. Let's multiply by $4$ and subtract $1$ each time:
$$2, 7, 27, 107, 427$$
The second to third jump is only off by one, so maybe ... (maybe!) ... we increase the number we subtract each time by one:
$$2, 7, 26, 101, 400$$
There it is.
If we apply the pattern again, then we get $4\times 400 - 5 = 1595.$
I just looked at the sequence and played with it.
$\endgroup$ 0 $\begingroup$$$\begin{align}2\times4-1&=7\\ 7\times4-2&=26\\ 26\times4-3&=101\\ 101\times4-4&=400\\ 400\times4-5&=1595\\\cdots\end{align}$$
By solving the linear recurrence $a_0=2,a_n=4a_{n-1}-n$,
$$a_n=\frac{14\cdot4^n+3n+4}9.$$
The general solution of the homogenous equation is
$$a_n=C4^n.$$
Searching a particular solution of the non-homogeneous equation, we try
$$a_n=An+B,$$
which gives
$$An+B=4(A(n-1)+B)-n,\\A=\frac13,B=\frac49.$$
Then
$$2=C4^0+\frac49,\\C=\frac{14}9.$$
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