The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule
. The next number would then be fourth power of 7 plus 1, or 2402.
And the harder way: Denote the <em>n</em>-th term in this sequence by
, and denote the given sequence by
.
Let
denote the <em>n</em>-th term in the sequence of forward differences of
, defined by
![b_n=a_{n+1}-a_n](https://tex.z-dn.net/?f=b_n%3Da_%7Bn%2B1%7D-a_n)
for <em>n</em> ≥ 1. That is,
is the sequence with
![b_1=a_2-a_1=17-2=15](https://tex.z-dn.net/?f=b_1%3Da_2-a_1%3D17-2%3D15)
![b_2=a_3-a_2=82-17=65](https://tex.z-dn.net/?f=b_2%3Da_3-a_2%3D82-17%3D65)
![b_3=a_4-a_3=175](https://tex.z-dn.net/?f=b_3%3Da_4-a_3%3D175)
![b_4=a_5-a_4=369](https://tex.z-dn.net/?f=b_4%3Da_5-a_4%3D369)
![b_5=a_6-a_5=671](https://tex.z-dn.net/?f=b_5%3Da_6-a_5%3D671)
and so on.
Next, let
denote the <em>n</em>-th term of the differences of
, i.e. for <em>n</em> ≥ 1,
![c_n=b_{n+1}-b_n](https://tex.z-dn.net/?f=c_n%3Db_%7Bn%2B1%7D-b_n)
so that
![c_1=b_2-b_1=65-15=50](https://tex.z-dn.net/?f=c_1%3Db_2-b_1%3D65-15%3D50)
![c_2=110](https://tex.z-dn.net/?f=c_2%3D110)
![c_3=194](https://tex.z-dn.net/?f=c_3%3D194)
![c_4=302](https://tex.z-dn.net/?f=c_4%3D302)
etc.
Again: let
denote the <em>n</em>-th difference of
:
![d_n=c_{n+1}-c_n](https://tex.z-dn.net/?f=d_n%3Dc_%7Bn%2B1%7D-c_n)
![d_1=c_2-c_1=60](https://tex.z-dn.net/?f=d_1%3Dc_2-c_1%3D60)
![d_2=84](https://tex.z-dn.net/?f=d_2%3D84)
![d_3=108](https://tex.z-dn.net/?f=d_3%3D108)
etc.
One more time: let
denote the <em>n</em>-th difference of
:
![e_n=d_{n+1}-d_n](https://tex.z-dn.net/?f=e_n%3Dd_%7Bn%2B1%7D-d_n)
![e_1=d_2-d_1=24](https://tex.z-dn.net/?f=e_1%3Dd_2-d_1%3D24)
![e_2=24](https://tex.z-dn.net/?f=e_2%3D24)
etc.
The fact that these last differences are constant is a good sign that
for all <em>n</em> ≥ 1. Assuming this, we would see that
is an arithmetic sequence given recursively by
![\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dd_1%3D60%5C%5Cd_%7Bn%2B1%7D%3Dd_n%2B24%26%5Ctext%7Bfor%20%7Dn%3E1%5Cend%7Bcases%7D)
and we can easily find the explicit rule:
![d_2=d_1+24](https://tex.z-dn.net/?f=d_2%3Dd_1%2B24)
![d_3=d_2+24=d_1+24\cdot2](https://tex.z-dn.net/?f=d_3%3Dd_2%2B24%3Dd_1%2B24%5Ccdot2)
![d_4=d_3+24=d_1+24\cdot3](https://tex.z-dn.net/?f=d_4%3Dd_3%2B24%3Dd_1%2B24%5Ccdot3)
and so on, up to
![d_n=d_1+24(n-1)](https://tex.z-dn.net/?f=d_n%3Dd_1%2B24%28n-1%29)
![d_n=24n+36](https://tex.z-dn.net/?f=d_n%3D24n%2B36)
Use the same strategy to find a closed form for
, then for
, and finally
.
![\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dc_1%3D50%5C%5Cc_%7Bn%2B1%7D%3Dc_n%2B24n%2B36%26%5Ctext%7Bfor%20%7Dn%3E1%5Cend%7Bcases%7D)
![c_2=c_1+24\cdot1+36](https://tex.z-dn.net/?f=c_2%3Dc_1%2B24%5Ccdot1%2B36)
![c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2](https://tex.z-dn.net/?f=c_3%3Dc_2%2B24%5Ccdot2%2B36%3Dc_1%2B24%281%2B2%29%2B36%5Ccdot2)
![c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3](https://tex.z-dn.net/?f=c_4%3Dc_3%2B24%5Ccdot3%2B36%3Dc_1%2B24%281%2B2%2B3%29%2B36%5Ccdot3)
and so on, up to
![c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)](https://tex.z-dn.net/?f=c_n%3Dc_1%2B24%281%2B2%2B3%2B%5Ccdots%2B%28n-1%29%29%2B36%28n-1%29)
Recall the formula for the sum of consecutive integers:
![1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2](https://tex.z-dn.net/?f=1%2B2%2B3%2B%5Ccdots%2Bn%3D%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5Enk%3D%5Cfrac%7Bn%28n%2B1%29%7D2)
![\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)](https://tex.z-dn.net/?f=%5Cimplies%20c_n%3Dc_1%2B%5Cdfrac%7B24%28n-1%29n%7D2%2B36%28n-1%29)
![\implies c_n=12n^2+24n+14](https://tex.z-dn.net/?f=%5Cimplies%20c_n%3D12n%5E2%2B24n%2B14)
![\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Db_1%3D15%5C%5Cb_%7Bn%2B1%7D%3Db_n%2B12n%5E2%2B24n%2B14%26%5Ctext%7Bfor%20%7Dn%3E1%5Cend%7Bcases%7D)
![b_2=b_1+12\cdot1^2+24\cdot1+14](https://tex.z-dn.net/?f=b_2%3Db_1%2B12%5Ccdot1%5E2%2B24%5Ccdot1%2B14)
![b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2](https://tex.z-dn.net/?f=b_3%3Db_2%2B12%5Ccdot2%5E2%2B24%5Ccdot2%2B14%3Db_1%2B12%281%5E2%2B2%5E2%29%2B24%281%2B2%29%2B14%5Ccdot2)
![b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3](https://tex.z-dn.net/?f=b_4%3Db_3%2B12%5Ccdot3%5E2%2B24%5Ccdot3%2B14%3Db_1%2B12%281%5E2%2B2%5E2%2B3%5E2%29%2B24%281%2B2%2B3%29%2B14%5Ccdot3)
and so on, up to
![b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)](https://tex.z-dn.net/?f=b_n%3Db_1%2B12%281%5E2%2B2%5E2%2B3%5E2%2B%5Ccdots%2B%28n-1%29%5E2%29%2B24%281%2B2%2B3%2B%5Ccdots%2B%28n-1%29%29%2B14%28n-1%29)
Recall the formula for the sum of squares of consecutive integers:
![1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6](https://tex.z-dn.net/?f=1%5E2%2B2%5E2%2B3%5E2%2B%5Ccdots%2Bn%5E2%3D%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5Enk%5E2%3D%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D6)
![\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)](https://tex.z-dn.net/?f=%5Cimplies%20b_n%3D15%2B%5Cdfrac%7B12%28n-1%29n%282%28n-1%29%2B1%29%7D6%2B%5Cdfrac%7B24%28n-1%29n%7D2%2B14%28n-1%29)
![\implies b_n=4n^3+6n^2+4n+1](https://tex.z-dn.net/?f=%5Cimplies%20b_n%3D4n%5E3%2B6n%5E2%2B4n%2B1)
![\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Da_1%3D2%5C%5Ca_%7Bn%2B1%7D%3Da_n%2B4n%5E3%2B6n%5E2%2B4n%2B1%26%5Ctext%7Bfor%20%7Dn%3E1%5Cend%7Bcases%7D)
![a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1](https://tex.z-dn.net/?f=a_2%3Da_1%2B4%5Ccdot1%5E3%2B6%5Ccdot1%5E2%2B4%5Ccdot1%2B1)
![a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2](https://tex.z-dn.net/?f=a_3%3Da_2%2B4%281%5E3%2B2%5E3%29%2B6%281%5E2%2B2%5E2%29%2B4%281%2B2%29%2B1%5Ccdot2)
![a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3](https://tex.z-dn.net/?f=a_4%3Da_3%2B4%281%5E3%2B2%5E3%2B3%5E3%29%2B6%281%5E2%2B2%5E2%2B3%5E2%29%2B4%281%2B2%2B3%29%2B1%5Ccdot3)
![\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1](https://tex.z-dn.net/?f=%5Cimplies%20a_n%3Da_1%2B4%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E3k%5E3%2B6%5Csum_%7Bk%3D1%7D%5E3k%5E2%2B4%5Csum_%7Bk%3D1%7D%5E3k%2B%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7D1)
![\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5Enk%5E3%3D%5Cfrac%7Bn%5E2%28n%2B1%29%5E2%7D4)
![\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)](https://tex.z-dn.net/?f=%5Cimplies%20a_n%3D2%2B%5Cdfrac%7B4%28n-1%29%5E2n%5E2%7D4%2B%5Cdfrac%7B6%28n-1%29n%282n%29%7D6%2B%5Cdfrac%7B4%28n-1%29n%7D2%2B%28n-1%29)
![\implies a_n=n^4+1](https://tex.z-dn.net/?f=%5Cimplies%20a_n%3Dn%5E4%2B1)