A closed form for a summation to find?

Question

I start as introduction with the well known formula for summing up natural numbers

$\sum _{k=1}^n k$= $\frac{1}{2} n (n+1)$ : formula from Gauss

proving this by induction $\sum _{k=1}^{n+1} k=\sum _{k=1}^n k+(n+1)$ = $\frac{1}{2} (n+2) (n+1)$(QED)

Another formulation as check
$\text{Assuming}\left[n\geq 1,\text{FullSimplify}\left[\sum _{k=1}^{n+1} k=\sum _{k=1}^n k+(n+1)\right]\right]$ = True

Assuming[n >= 1, FullSimplify[Sum[k, {k, 1, n + 1}] == Sum[k, {k, 1, n}] + (n + 1)]

$\sum _{k=1}^{n+1} k=\sum _{k=1}^n k+(n+1)$ (summing up natural numbers)(1,2,3,..n-1, n, n+1,..)

Simplify[Sum[k, {k, 1, n + 1}] == Sum[k, {k, 1, n}] + (n + 1)]

Now i go to a more difficult expression..

((-1)^i*(1 - i + k)^(4 + 2*k))/(i!*(2 - i + 2*k)!) = ((k + 2)*(2*k + 3)*(k + 1))/12

How can prove this by induction ..by using the idea of the simpler example ?

Answer 1

So, yes, in the second expression the mathematical formulation has to be properly written in the OP. The sum over $i$ goes from $0$ to $k$, in order for the expression to makes sense. I demonstrate below.

The expression we want to obtain:

exp2 = ((k + 2)*(2*k + 3)*(k + 1))/12

Do some values

Table[exp2, {k, 0, 10}]

which yields

{1/2, 5/2, 7, 15, 55/2, 91/2, 70, 102, 285/2, 385/2, 253}

So, now we go back to the LHS and we have the following values

Table[With[{k = xx}, 
  Sum[((-1)^i*(1 - i + k)^(4 + 2*k))/(i!*(2 - i + 2*k)!), {i, 0, 
    k}]], {xx, 0, 10}]

which gives

{1/2, 5/2, 7, 15, 55/2, 91/2, 70, 102, 285/2, 385/2, 253}

which is obviously the right thing to do.

To get the analytic formula

FindSequenceFunction[{1/2, 5/2, 7, 15, 55/2, 91/2, 70, 102, 285/2, 
    385/2, 253}, k] /. k -> k + 1 // FullSimplify

which returns

1/12 (1 + k) (2 + k) (3 + 2 k)

and now we are happy.