Respuesta :
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 [tex]n^4+1[/tex]. The next number would then be fourth power of 7 plus 1, or 2402.
And the harder way: Denote the n-th term in this sequence by [tex]a_n[/tex], and denote the given sequence by [tex]\{a_n\}_{n\ge1}[/tex].
Let [tex]b_n[/tex] denote the n-th term in the sequence of forward differences of [tex]\{a_n\}[/tex], defined by
[tex]b_n=a_{n+1}-a_n[/tex]
for n ≥ 1. That is, [tex]\{b_n\}[/tex] is the sequence with
[tex]b_1=a_2-a_1=17-2=15[/tex]
[tex]b_2=a_3-a_2=82-17=65[/tex]
[tex]b_3=a_4-a_3=175[/tex]
[tex]b_4=a_5-a_4=369[/tex]
[tex]b_5=a_6-a_5=671[/tex]
and so on.
Next, let [tex]c_n[/tex] denote the n-th term of the differences of [tex]\{b_n\}[/tex], i.e. for n ≥ 1,
[tex]c_n=b_{n+1}-b_n[/tex]
so that
[tex]c_1=b_2-b_1=65-15=50[/tex]
[tex]c_2=110[/tex]
[tex]c_3=194[/tex]
[tex]c_4=302[/tex]
etc.
Again: let [tex]d_n[/tex] denote the n-th difference of [tex]\{c_n\}[/tex]:
[tex]d_n=c_{n+1}-c_n[/tex]
[tex]d_1=c_2-c_1=60[/tex]
[tex]d_2=84[/tex]
[tex]d_3=108[/tex]
etc.
One more time: let [tex]e_n[/tex] denote the n-th difference of [tex]\{d_n\}[/tex]:
[tex]e_n=d_{n+1}-d_n[/tex]
[tex]e_1=d_2-d_1=24[/tex]
[tex]e_2=24[/tex]
etc.
The fact that these last differences are constant is a good sign that [tex]e_n=24[/tex] for all n ≥ 1. Assuming this, we would see that [tex]\{d_n\}[/tex] is an arithmetic sequence given recursively by
[tex]\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}[/tex]
and we can easily find the explicit rule:
[tex]d_2=d_1+24[/tex]
[tex]d_3=d_2+24=d_1+24\cdot2[/tex]
[tex]d_4=d_3+24=d_1+24\cdot3[/tex]
and so on, up to
[tex]d_n=d_1+24(n-1)[/tex]
[tex]d_n=24n+36[/tex]
Use the same strategy to find a closed form for [tex]\{c_n\}[/tex], then for [tex]\{b_n\}[/tex], and finally [tex]\{a_n\}[/tex].
[tex]\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}[/tex]
[tex]c_2=c_1+24\cdot1+36[/tex]
[tex]c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2[/tex]
[tex]c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3[/tex]
and so on, up to
[tex]c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)[/tex]
Recall the formula for the sum of consecutive integers:
[tex]1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2[/tex]
[tex]\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)[/tex]
[tex]\implies c_n=12n^2+24n+14[/tex]
[tex]\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}[/tex]
[tex]b_2=b_1+12\cdot1^2+24\cdot1+14[/tex]
[tex]b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2[/tex]
[tex]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[/tex]
and so on, up to
[tex]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)[/tex]
Recall the formula for the sum of squares of consecutive integers:
[tex]1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6[/tex]
[tex]\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)[/tex]
[tex]\implies b_n=4n^3+6n^2+4n+1[/tex]
[tex]\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}[/tex]
[tex]a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1[/tex]
[tex]a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2[/tex]
[tex]a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3[/tex]
[tex]\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[/tex]
[tex]\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4[/tex]
[tex]\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)[/tex]
[tex]\implies a_n=n^4+1[/tex]
