What is the nth term of the sequence2,6,12,20,?

Question

What is the nth term of the sequence2,6,12,20,?

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Arya 1 month 2021-08-14T08:12:13+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-08-14T08:13:53+00:00

    Given the sequence

    2

    ,

    6

    ,

    12

    ,

    20

    , write it out in a line with gaps between each term:

    2

    00000

    6

    00000

    12

    00000

    20

    Add a line of term below listing the differences between each pair of terms:

    2

    00000

    6

    00000

    12

    00000

    20

    000

    4

    00000

    6

    000000

    8

    Add another line of term below listing the differences between each pair of terms:

    2

    00000

    6

    00000

    12

    00000

    20

    000

    4

    00000

    6

    000000

    8

    000000

    2

    000000

    2

    Notice that both terms of the last line are the same.

    That implies that our given sequence can be matched using a quadratic formula, which we can construct using the first term of each of the lines as a coefficient:

    a

    n

    =

    2

    0

    !

    +

    4

    1

    !

    (

    n

    1

    )

    +

    2

    2

    !

    (

    n

    1

    )

    (

    n

    2

    )

    a

    n

    =

    2

    +

    4

    n

    4

    +

    n

    2

    3

    n

    +

    2

    a

    n

    =

    n

    2

    +

    n

    If you know the formulas for

    N

    n

    =

    1

    n

    2

    and

    N

    n

    =

    1

    n

    then you can just add them to provide the formula for

    N

    n

    =

    1

    a

    n

    .

    I would rather construct it directly:

    Add another line to the top of our sequences, consisting of the sums of the terms on the original top line:

    0

    00000

    2

    00000

    8

    00000

    20

    00000

    40

    000

    2

    00000

    6

    00000

    12

    00000

    20

    000000

    4

    00000

    6

    000000

    8

    000000000

    2

    000000

    2

    We want to start with

    s

    1

    =

    2

    . So ignore the first terms of each sequence and use the second terms to give us our formula for the sum to

    n

    terms:

    s

    n

    =

    2

    0

    !

    +

    6

    1

    !

    (

    n

    1

    )

    +

    6

    2

    !

    (

    n

    1

    )

    (

    n

    2

    )

    +

    2

    3

    !

    (

    n

    1

    )

    (

    n

    2

    )

    (

    n

    3

    )

    s

    n

    =

    2

    +

    6

    n

    6

    +

    3

    n

    2

    9

    n

    +

    6

    +

    1

    3

    n

    3

    2

    n

    2

    +

    11

    3

    n

    2

    s

    n

    =

    1

    3

    (

    n

    3

    +

    3

    n

    2

    +

    2

    n

    )

    s

    n

    =

    1

    3

    n

    (

    n

    +

    1

    )

    (

    n

    +

    2

    )

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