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 ?
((-1)^i*(1 - i + k)^(4 + 2*k))/(i!*(2 - i + 2*k)!) = ((k + 2)*(2*k + 3)*(k + 1))/12
should be written with a==
, not a=
, and it's false. Perhaps aSum
or two is missing. $\endgroup$