How to find the nth term of quadratic sequences

Question

1,3,7,13,21

I got this question in my exam today, and it baffled me. When I got home I tried to solve it, but I kept getting the wrong answer. If you know how to solve this, please help me out by telling me how you solved it. Thanks in advance.

Answer 1

HINT: You are given that it is quadratic, so it is in the form $$f(x)=ax^2+bx+c$$ You are also given $$f(1)=1$$ $$f(2)=3$$ $$f(3)=7$$ $$f(4)=13$$ $$f(5)=21$$ You can use this to say that $$a(1)^2+b(1)+c=1$$ $$a(2)^2+b(2)+c=3$$ $$a(3)^2+b(3)+c=7$$ or $$a+b+c=1$$ $$4a+2b+c=3$$ $$9a+3b+c=7$$ Can you use this to solve for the coefficients of the quadratic?

Answer 2

$$\begin{array} &&&&&&&\\ r=&1 &&2&&3&&4&&5&&\cdots &n\\\\ T_r=&1 & &3 &&7 &&13 &&21&&\cdots &T_n\\ \Delta T_r=&&2 && 4&& 6 && 8\\ \Delta^2 T_r=&&& 2 &&2 &&2 & \end{array}$$

From the above it is clear that $$\color{red}{T_n}=1+\sum_{i=1}^{n-1}2i\color{red}{=n^2-n+1}$$

Answer 3

Note: $$a_2=a_1+2;$$ $$a_3=a_2+4=a_1+2+4;$$ $$a_4=a_3+6=a_1+2+4+6;$$ $$\cdots$$ $$a_n=a_1+2+4+\cdots+2(n-1)=1+2(1+2+\cdots+(n-1))=1+2\cdot \frac{1+n-1}{2}\cdot (n-1)=n^2-n+1.$$

Answer 4

A good method for this and harder sequences is to match each term of the sequence with its corresponding $n$, then take the differences $\delta_1$ of the terms of the sequence, then the differences of the differences $\delta_2$, until a constant is reached:

$$n = 1, 2, 3, 4, 5...$$ $$a_n = 1, 3, 7, 13, 21...$$ $$\delta_1 = 2,4,6,8...$$ $$\delta_2 = 2,2,2...$$ Beginning with the $\delta_2$ sequence, extrapolate backward until each sequence has $n+1$ terms:$$n = 0,1,2,3,4,5...$$ $$a_n = 1,1,3,7,13,21...$$ $$\delta_1 = -2,0, 2,4,6,8...$$ $$\delta_2 = 2,2,2,2,2,2...$$ $\delta_2$ is constant. We see that $$\delta_1 = \sum2-2 = 2n - 2$$ Note that $-2$ is $\delta_1$ for $n=0$. Finally, $$a_n = \sum(2n-2)+1$$ $$=2\sum{n}-2n+1$$ $$=2\frac{(n^2+n)} {2}-2n+1$$ $$=n^2-n+1$$ Again, the $+1$ is $a_n$ for $n=0$. This method of differences works well for sequences higher than the quadratic, but of course then requires summation not just of consecutive integers, as above, but of the squares, cubes, etc. of consecutive integers.