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This seems easy....

Pattern:

3, 6, 10, 15, 21....

How do I find expression for this?

It looked easy, but I couldn't find it.

I tried graphing it, but it turned out that it is not quadratic...

Thanks.

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    $\begingroup$ It is quadratic, $f(n)=(n+1)(n+2)/2$ for $n=1,2,\dots$. Perhaps you recognize $1+2$, $1+2+3$, $1+2+3+4$, and so on. $\endgroup$ – André Nicolas Dec 2 '14 at 16:37
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    $\begingroup$ $6=3+3,10=6+4,15=10+5,21=15+6,a_{n+1}=a_n+n+2$ $\endgroup$ – mike Dec 2 '14 at 16:39
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$f(n)=\binom{n+2}{2}=\frac{(n+1)(n+2)}{2}$

If you have a sequence where the differences between terms is lineal then it is going to be a quadratic polynomial.

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$$3\quad 3+\color{red}3=6\quad 6+\color{red}4=10\quad 10+\color{red}5=15\quad 15+\color{red}6=21\quad $$

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If you don't recognize the pattern (Triangular numbers), you can look at the differences.

They are 3,4,5,6. Since the differences are an arithmetic sequence, you know the series is a quadratic, so can start with $n^2$.

Then, you can call your expression $an^2+bn+c$ and solve it by using 3 of the values from your sequence.

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$$~~~~~~3 ~~~~~~ 6 ~~~~~~ 10 ~~~~~~ 15 ~~~~~~ 21 \\3 ~~~~~~~~ 4 ~~~~~~~~~ 5 ~~~~~~ 6 \\~1 ~~~~~~~~ 1 ~~~~~~~ 1$$

So $f(x) = ax^2 + bx + c$ Now $f(1)=3,f(2)=6,f(3)=10.$ From this we can solve $a,b,c$

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  • $\begingroup$ How do we exactly do that? $\endgroup$ – didgocks Dec 2 '14 at 16:48
  • $\begingroup$ @didgocks: Take the differences of the numbers in the first step. Then consequently in the next step. when you see all the differences are equal stop and see how many differences you have taken that is your order of the pattern. $\endgroup$ – Argha Dec 2 '14 at 16:53
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If you prefer recursion: $a_n=a_{n-1}+n-1+3$ with $a_1=3$.

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