Agohofidov6

2022-01-04

Determine the general term, $t}_{n$ , for the following sequence.

$8,20,36,56$

Bob Huerta

Beginner2022-01-05Added 41 answers

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)

$4a+2b+c-(a+b+c)=20-8$

$3a+b=12$ (4)

Now, subtract equation (2) from (3)

$9a+3b+c-(4a+2b+c)=36-20$

$5a+b=16$ (5)

So,

$5a+b-(3a+b)=16-12$

$2a=4$

$a=2$

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

After the second term the difference is the same, so n-th term of this series would be polinomial with degree 2.

So,

but

Solve all the equations. Subtract equation (1) from (2)

Now, subtract equation (2) from (3)

So,

Calculate b

From equation (1) find c

Consider n-th term

and

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