Agohofidov6

2022-01-04

Determine the general term, ${t}_{n}$, for the following sequence.
$8,20,36,56$

Bob Huerta

Check the difference in each term
After the second term the difference is the same, so n-th term of this series would be polinomial with degree 2.
So, ${t}_{n}=a{n}^{2}+bn+c$
${t}_{1}=8$, ${t}_{2}=20$, ${t}_{3}=36$
but $n=1$, ${t}_{1}=8$ as we have below
$a\left(1\right)+b\left(1\right)+c=8$
$a+b+c=8$ (1)
${t}_{2}=20$, $n=2$, so we have
$a{\left(2\right)}^{2}+b\left(2\right)+c=20$
$4a+2b+c=20$ (2)
${t}_{3}=36$, $n=3$
$a{\left(3\right)}^{2}+b\left(3\right)+c=36$
$9a+3b+c=36$ (3)
Solve all the equations. Subtract equation (1) from (2)
$4a+2b+c-\left(a+b+c\right)=20-8$
$3a+b=12$ (4)
Now, subtract equation (2) from (3)
$9a+3b+c-\left(4a+2b+c\right)=36-20$
$5a+b=16$ (5)
So,
$5a+b-\left(3a+b\right)=16-12$
$2a=4$
$a=2$
Calculate b
$5\left(2\right)+b=16$
$b=6$
From equation (1) find c
$2+6+c=8$
$c=0$
Consider n-th term
${t}_{n}=a{n}^{2}+bn+c$
and $a=2$, $b=6$, $c=0$
${t}_{n}=\left(2\right)\left({n}^{2}\right)+6\left(n\right)+0$
${t}_{n}=2{n}^{2}+6n$ is the answer

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