# How to solve this geometric/arithmetic sequence problem without guessing and checking?

Which represents the sum of $6+15+26+39$ ?

1. $\sum\limits_{n=4}^7 (2n-2)$

2. $\sum\limits_{n=4}^7 (n^2-10)$

3. $\sum\limits_{n=2}^5 (n^2+n)$

4. $\sum\limits_{n=2}^5 \mathopen{\Big(}\dfrac{3n^2}{2}\mathclose{\Big)}$

I can employ guess and check and arrive at the answer of choice 2 (by just plugging in each variable, but is there any way to directly solve for the equation (ie, not given the answer choices)?

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## migrated from mathematica.stackexchange.comApr 5 '13 at 2:50

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I don't know what you mean by "directly solve for the equation" (at any rate, you mean expression, not equation; note the lack of equals signs), because you can find infinitely many expressions that will have the same value as $$6+15+26+39=86.$$ You can even find infinitely many polynomials $f$ and integers $a$ such that $$f(a)=6,\quad f(a+1)=15,\quad f(a+2)=26,\quad f(a+3)=39$$ so that the expression $$\sum_{n=a}^{a+3}f(n)$$ represents the same summation. There is no "canonical" or "natural" way of taking a sum of integers and making an expression that "does the same thing". In short: the only way of solving the question you are considering is to check the answer choices given.

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Here's one possible answer, using Mathematica's built in function:

FindSequenceFunction[{6, 15, 26, 39}]


-1 + 6 #1 + #1^2 &

x^2 + 6 x -1