Question

How to find nth term for a sequence of following numbers;

1,9,36,100...? I know that this is square sequence and I use the formula a+(n-1)d+1/2×(n-1)×(n-2)c where a=the first term of first sequence d=difference between first two terms in first sequence and c=common difference between terms in the second difference. How to express c in this example if the terms in the second sequence are cube numbers? please help

Answer 1

Let
`a_n` denote the given sequence.
`a_n` has forward differences

{9 - 1, 36 - 9, 100 - 36, ...} = {8, 27, 64, ...} = {2^3, 3^3, 4^3, ...}

If we call the sequence of forward differences
`b_n`, then for
`n\ge1`,

`b_n=(n+1)^3`

`b_n` is defined in terms of
`a_n` for all
`n\ge1` by

`b_n=a_(n+1)-a_n`

and so
`a_n` is defined recursively by

`a_n=\begin{cases}a_1=1\\a_(n+1)=a_n+(n+1)^3&\text{for }n\ge1\end{cases}`

We can deduce a pattern for the general
`n`-th term:

`a_2=a_1+2^3`

`a_3=a_2+3^3=a_1+\displaystyle\sum_(i=1)^2(i+1)^3`

`a_4=a_3+4^3=a_1+\displaystyle\sum_(i=1)^3(i+1)^3`

and so on, up to

`a_n=a_1+\displaystyle\sum_(i=1)^(n-1)(i+1)^3`

We can simplify the right hand side a bit, noticing that
`a_1=1=1^3` matches
`(i+1)^3` for
`i=0`:

`a_n=\displaystyle\sum_(i=0)^(n-1)(i+1)^3`

and to simplify things a bit more, we shift the index of summation:

`a_n=\displaystyle\sum_(i=1)^ni^3`

You should know that the right side has a nice closed form (look up "Faulhaber's formula" if you don't):

`a_n=\frac{n^2(n+1)^2}4`