Jacob Bernoulli’s power sum problem

For any positive integer p, find


“The following elegant solution [says Dörrie] is based upon the binomial theorem.
By resorting to the device of considering the magnitudes B^1,B^2,B^3,\dots ,B^v resulting from the binomial expansion of (x+B)^v as unknowns subject to v certain conditions rather than as powers of B, we obtain an amazingly short derivation of S.”

(This type of crazy-looking procedure is Blissard’s symbolic method, or umbral calculus, introduced by John Blissard in 1861. Also used extensively by Lucas. Now referred to as 19th C or classic umbral calculus, as since the 1970s the subject has been transformed by Roman, Rota and others.)

According to the binomial theorem, if P=p+1,

    \[ (v+B)^P=v^P+Pv^pB^1+\tbinom{P}{2}v^{p-1}B^2+\tbinom{P}{3}v^{p-2}B^3\dots\\ \]


    \begin{align*} (v+B-1)^P&=v^P+Pv^p(B-1)^1+\tbinom{P}{2}v^{p-1}(B-2)^2\\ &+\tbinom{P}{3}v^{p-2}(B-1)^3\dots\\ \end{align*}

Since B^1-(B^1-1)=1,  subtracting the second equation from the first gives:

    \begin{align*} \mathrm{(I)} \qquad (v+B)^P-(v+B-1)^P&=Pv^p+\tbinom{P}{2}v^{p-1}[B^2-(B-2)^2]\\ &+\tbinom{P}{3}v^{p-2}[B^3-(B-1)^3]\dots \end{align*}

Now define the unknowns B^1,B^2,B^3,\dots by the equations

(1) \ (B-1)^2=B^2, \kern 1pc (2) \ (B-1)^3=B^3, \kern 1pc (3) \ (B-1)^4=B^4, \kern 1pc etc.

This simplifies \mathrm{(I)} to

    \[ Pv^p=(v+B)^P-(v-1+B)^P \]

Substitute in v=1,2,3,\dots,n to get

    \begin{align*} P\cdot 1^p&=(1+B)^P-B^P,\\ P\cdot 2^p&=(2+B)^P-(1+B)^P,\\ P\cdot 3^p&=(3+B)^P-(2+B)^P,\\ &\vdots\\ P\cdot n^p&=(n+B)^P-(n-1+B)^P.\\ \end{align*}

Addition of these n equations gives

    \[ \mathrm{(II)} \qquad PS=(n+B)^P-B^P \]


    \[ \mathrm{(IIa)} \qquad 1^p+2^p+3^p+4^p+...+n^p=\frac{(n+B)^P-B^P}{P} \]

Now to determine B^1, B^2, B^3\dots from equations (1), (2), (3)\dots

From (1) it follows that B^2-2B^1+1=B^2 \ \to \ -2B^1+1=0 \ \to \ B^1=1/2
From (2), -3B^2+3B^1-1=0 \ \to \ -3B^2+3/2=1 \to \ B^2=1/6
From (3), -4B^3+6B^2-4B^1+1=0
\hspace{3cm} \to \ -4B^3+1-2+1=0 \ \to \ B^3=0
From (4), B^4=-1/30

These are known as the Bernoulli numbers:

B_0 = 1, B_1 = \pm 1/2, B_2 = 1/6, B_3 = 0, B_4 = -1/30, B_5 = 0, B_6 = 1/42, B_7 = 0, B_8 = -1/30\dots

Then from \mathrm{(IIa)} we get

    \begin{align*} 1+2+3+\dots +n&=\frac{(n+B)^2-B^2}{2}=\frac{n^2+2nB^1}{2}\\ &=n\frac{n+1}{2} \end{align*}

    \begin{align*} 1^2+2^2+3^2+\dots +n^2&=\frac{(n+B)^3-B^3}{3}=\frac{n^3+3n^2B^1+3nB^2}{3}\\ &=\frac{1}{6}n(n+1)(2n+1) \end{align*}

    \begin{align*} 1^3+2^3+3^3+\dots +n^3&=\frac{(n+B)^4-B^4}{4}\\ &=\frac{n^4+4n^3B^1+6n^2B^2+4nB^3}{4}\\ &=\left(n\frac{n+1}{2}\right)^2 \end{align*}

    \begin{align*} 1^4+2^4+3^4+\dots +n^4&=\frac{(n+B)^5-B^5}{5}\\ &=\frac{n^5+5n^4B^1+10n^3B^2+10n^2B^3+5nB^4}{5}\\ &=\frac{n(n+1)(2n+1)(3n(n+1)-1)}{30} \end{align*}

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