Sum the k power

Question

Please find the closed form for the sum of the first $n$ positive integers to the $5^{th}$ power(or, the first n positive integers to the power 5) in at least 3 different ways.

Notation

Sequence $\{S_n=\sum_{k=1}^{n}k^5\}$ refers the sum of the first $n$ positive integers to the $5^{th}$ power.

Sequence $\{T_n=\sum_{k=1}^{n}k^6\}$ refers the sum of the first $n$ positive integers to the $6^{th}$ power.

Solution One(Guess and Prove)

Guess: $S_n=\frac{1}{6}n^6+\frac{1}{2}n^5+\frac{5}{12}n^4-\frac{1}{12}n^2,\ \ for \ \ n\geq1$

We guess the closed form for the sum of the first $n$ positive integers to the $5^{th}$ power would be the same as that of the $2^{th}$ power, which remains in polynomial form. Then we use the method of undetermined coefficients to get the closed form of $S_n$.

Assume $S_n=a_6n^6+a_5n^5+a_4n^4+a_3n^3+a_2n^2+a_1n+a_0$, and calculate the first seven value of $S_n$.

$$\begin{aligned} &S_1=1=a_0+a_1+a_2+a_3+a_4+a_5+a_6\\ &S_2=33=a_0+2a_1+4a_2+8a_3+16a_4+32a_5+64a_6\\ &S_3=276=a_0+3a_1+9a_2+27a_3+81a_4+243a_5+729a_6\\ &S_4=1300=a_0+4a_1+16a_2+64a_3+256a_4+1024a_5+4096a_6\\ &S_5=4425=a_0+5a_1+25a_2+125a_3+625a_4+3125a_5+15625a_6\\ &S_6=12201=a_0+6a_1+36a_2+216a_3+1296a_4+7776a_5+46656a_6\\ &S_7=29008=a_0+7a_1+49a_2+343a_3+2401a_4+16807a_5+117649a_6 \end{aligned}$$

$N=\left[\begin{matrix}1 & 1 & 1 & 1 & 1 & 1 & 1\\1 & 2 & 4 & 8 & 16 & 32 & 64\\1 & 3 & 9 & 27 & 81 & 243 & 729\\1 & 4 & 16 & 64 & 256 & 1024 & 4096\\1 & 5 & 25 & 125 & 625 & 3125 & 15625\\1 & 6 & 36 & 216 & 1296 & 7776 & 46656\\1 & 7 & 49 & 343 & 2401 & 16807 & 117649\end{matrix}\right]$, $S=\left[\begin{matrix}1\\33\\276\\1300\\4425\\12201\\29008\end{matrix}\right]$,coefficient $C=\left[\begin{matrix}a_0\\a_1\\a_2\\a_3\\a_4\\a_5\\a_6\end{matrix}\right]$,

We can solve the equation $NC=S$ and get the solution $C=N^{-1}S=\left[\begin{matrix}0,0,- \frac{1}{12},0,\frac{5}{12},\frac{1}{2},\frac{1}{6}\end{matrix}\right]^T$.

Finally, we guess $S_n=\frac{1}{6}n^6+\frac{1}{2}n^5+\frac{5}{12}n^4-\frac{1}{12}n^2,\ \ for \ \ n\geq1$.

Prove: When $n=1$, $S_1=1=\frac{1}{6}1^6+\frac{1}{2}1^5+\frac{5}{12}1^4-\frac{1}{12}1^2=\frac{1}{6}+\frac{1}{2}+\frac{5}{12}-\frac{1}{12}$, so the basis is checked. For the induction, suppose that $n>1$, and assume the formula mentioned above holds when $n$ is replaced by $n-1$. Since $S_n=S_{n-1}+n^5$, we have

$ \begin{aligned} & S_n\ =\frac{1}{6}(n-1)^6+\frac{1}{2}(n-1)^5+\frac{5}{12}(n-1)^4-\frac{1}{12}(n-1)^2+n^5 \\ &\ \ \ \ \ =\frac{1}{6}(n^6-6n^5+15n^4-20n^3+15n^2-6n+1)+\frac{1}{2}(n^5-5n^4+10n^3-10n^2+5n-1)\\ &\ \ \ \ \ \ \ \ \ \ +\frac{5}{12}(n^4-4n^3+6n^2-4n+1)-\frac{1}{12}(n^2-2n+1)+n^5 \\&\ \ \ \ \ =\frac{1}{6}n^6+\frac{1}{2}n^5+\frac{5}{12}n^4-\frac{1}{12}n^2 \end{aligned} $

Therefore, the formula indeed holds beyond a reasonable doubt for all $n\geq 1$.

Solution Two(Perturb the sum)

We start with the sum of the first $n$ positive integers to the $6^{th}$ power $T_n$.

$T_n+(n+1)^6=\sum_{k=1}^{n+1}k^6=\sum_{k=0}^{n}(k+1)^6=\sum_{k=0}^{n}(k^6+6k^5+15k^4+20k^3+15k^2+6k+1)\\\begin{aligned}&=\sum_{k=0}^{n}k^6+6\sum_{k=0}^{n}k^5+15\sum_{k=0}^{n}k^4+20\sum_{k=0}^{n}k^3+15\sum_{k=0}^{n}k^2+6\sum_{k=0}^{n}k+n+1\\&=T_n+6S_n+15\sum_{k=0}^{n}k^4+20\sum_{k=0}^{n}k^3+15\sum_{k=0}^{n}k^2+6\sum_{k=0}^{n}k+n+1\end{aligned}$

As we know,

$\begin{aligned}&\sum_{k=0}^{n}k=\frac{n(n+1)}{2}\\&\sum_{k=0}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\end{aligned}$

Then we can perturb the sum again to solve $\sum_{k=0}^{n}k^4$ and $\sum_{k=0}^{n}k^3$ :

$ \sum_{k=1}^{n}k^4+(n+1)^4=\sum_{k=1}^{n+1}k^4=\sum_{k=0}^{n}(k+1)^4=\sum_{k=0}^{n}(k^4+4k^3+6k^2+4k+1)\\\begin{aligned}&=\sum_{k=0}^{n}k^4+4\sum_{k=0}^{n}k^3+6\sum_{k=0}^{n}k^2+4\sum_{k=0}^{n}k+n+1\end{aligned}$

We can get $\sum_{k=0}^{n}k^3=\frac{1}{4}((n+1)^4-6\sum_{k=0}^{n}k^2-4\sum_{k=0}^{n}k-n-1)=\frac{1}{4}(n^4+2n^3+n^2)$ from the formula above.

$ \sum_{k=1}^{n}k^5+(n+1)^5=\sum_{k=1}^{n+1}k^5=\sum_{k=0}^{n}(k+1)^5=\sum_{k=0}^{n}(k^5+5k^4+10k^3+10k^2+5k+1)\\\begin{aligned}&=\sum_{k=0}^{n}k^5+5\sum_{k=0}^{n}k^4+10\sum_{k=0}^{n}k^3+10\sum_{k=0}^{n}k^2+5\sum_{k=0}^{n}k+n+1\end{aligned}$

We can get

$\sum_{k=0}^{n}k^4=\frac{1}{5}((n+1)^5-10\sum_{k=0}^{n}k^3-10\sum_{k=0}^{n}k^2-5\sum_{k=0}^{n}k-n-1)=\frac{1}{5}n^5+\frac{1}{2}n^4+\frac{1}{3}n^3-\frac{1}{30}n$

from the formula above.

Finally, we can get

$S_n=\frac{1}{6}((n+1)^6-15\sum_{k=0}^{n}k^4-20\sum_{k=0}^{n}k^3-15\sum_{k=0}^{n}k^2-6\sum_{k=0}^{n}k-n-1)=\frac{1}{6}n^6+\frac{1}{2}n^5+\frac{5}{12}n^4-\frac{1}{12}n^2$

which is the same as the answer solved by the first solution.

Solution Three(Replace sums by integrals)

First, we know that $S_n$ is approximately $\int_0^nx^5dx=\frac{1}{6}n^6$. Then we consider the error in the approximation, $E_n=S_n-\frac{1}{6}n^6$. Since $S_n$ satisfies the recurrence $S_n=S_{n-1}+n^5$, we find that $E_n$ satisfies the recurrence

$$\begin{aligned}E_n&=S_n-\frac{1}{6}n^6\\&=S_{n-1}+n^5-\frac{1}{6}n^6\\&=E_{n-1}+\frac{1}{6}(n-1)^6+n^5-\frac{1}{6}n^6\\&=E_{n-1}+\frac{5}{2}n^4-\frac{10}{3}n^3+\frac{5}{2}n^2-n+\frac{1}{6}\\&=E_0+\frac{5}{2}\sum_{k=1}^nk^4-\frac{10}{3}\sum_{k=1}^nk^3+\frac{5}{2}\sum_{k=1}^nk^2-\sum_{k=1}^nk+\frac{n}{6}\end{aligned}$$

It’s easy to check $E_0=0$.

Similarly, we can get $\sum_{k=1}^nk^4$ and $\sum_{k=1}^nk^3$ by replacing sums by integrals as well (they are sub-problems). But we use the result we have got in the solution two,

$$\begin{aligned}&\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\\&\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\\&\sum_{k=1}^{n}k^3=\frac{1}{4}(n^4+2n^3+n^2)\\&\sum_{k=1}^{n}k^4=\frac{1}{5}n^5+\frac{1}{2}n^4+\frac{1}{3}n^3-\frac{1}{30}n\end{aligned}$$

Therefore, we find $E_n=\frac{1}{2}n^5+\frac{5}{12}n^4-\frac{1}{12}n^2$, and $S_n=E_n+\frac{1}{6}n^6=\frac{1}{6}n^6+\frac{1}{2}n^5+\frac{5}{12}n^4-\frac{1}{12}n^2$.