Vtyjkyuk

I started to read some book on elementary number theory and preliminary chapter asks to establish some formulas by mathematical induction:

There is this formula:

$$1+2+...+n= dfrac n(n+1)2$$

And this one:

$$ 1cdot 2 + 2cdot 3 + ... + ncdot (n+1)= dfrac n(n+1)(n+2)3$$

With some pencil-and-paper work I established that this should hold:

$$1cdot 2 cdot ... cdot m + 2cdot 3 cdot ... cdot (m+1) + ... + n cdot (n+1) cdot ... cdot (n+m-1)= dfrac n(n+1)...(n+m-1)(n+m)m+1$$

I did not prove this formula that I established but just checked some cases and it seems to hold.

It can be written in the form of a hockey-stick identity, I think, so it holds.

Now, I know about generalization of first formula that goes like this (Faulhaber´s formula):

$$1^w + 2^w + ... + n^w=dfrac 1w+1 cdot displaystyle sum_j=0^w w+1 choose j B_j n^w+1-j$$

How do the generalization $$(1 cdot 2 cdot ... cdot m)^w+ (2 cdot 3 cdot ... cdot (m+1))^w+...+(n cdot(n+1)+ cdot ... cdot (n+m-1))^w=?$$ look like? That is, what is on the right side?

I think there is no simple closed form. Using your notation, if we denote

$$w=a_1$$

$$sumlimits_q=1^m a_q=b_m$$

so

$$sumlimits_k=1^n left(fracm+k-1k-1right)^w=sumlimits_s=1^m-1sumlimits_a_s+1=0^a_sfraca_1!a_m!prodlimits_t=1^m-1frac1(a_t-a_t+1)!left(left[a_1atop a_m-t+1right]right)^a_t-a_t+1sumlimits_k=1^nk^b_m$$

which is a little bit ugly.

For your first conjecture:

Write

$1cdot 2 cdot ... cdot m + 2cdot 3 cdot ... cdot (m+1) + ... + n cdot (n+1) cdot ... cdot (n+m-1)= dfrac n(n+1)...(n+m-1)(n+m)m+1
$

in the form
$s(n, m)
=t(n, m)
$
where
$s(n, m)
=sum_k=1^n prod_j=k^k+m-1 j
$
and
$t(n, m)
=dfracprod_j=n^n+mjm+1
$.

I will prove this
by induction on $n$.

For $n=1$
this is
$=prod_j=1^m j
=dfracprod_j=1^1+mjm+1
$
or
$m! = dfrac(m+1)!m+1
$
so that
$s(1, m) = t(1, m)$.

Consider

$beginarray\
s(n+1,m)-s(n,m)
&=sum_k=1^n+1 prod_j=k^k+m-1 j-sum_k=1^n prod_j=k^k+m-1 j\
&=prod_j=n+1^n+m j\
endarray
$

and

$beginarray\
t(n+1, m)-t(n, m)
&=dfracprod_j=n+1^n+1+mjm+1-dfracprod_j=n^n+mjm+1\
&=dfracprod_j=n+1^n+1+mj-prod_j=n^n+mjm+1\
&=dfracprod_j=n+1^n+mjleft((n+1+m)-nright)m+1\
&=dfracprod_j=n+1^n+mjleft(m+1right)m+1\
&=prod_j=n+1^n+mj\
&=s(n+1,m)-s(n,m)\
endarray
$

Since
$s(1, m) = t(1, m)$,
this shows that
$s(n, m) = t(n, m)$.