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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 ?

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  • $\begingroup$ The LHS of the second expression that you want to prove has an $i$ and $k$. Is $i$ summed over a domain? $\endgroup$
    – user49048
    Commented Feb 23, 2022 at 14:10
  • $\begingroup$ (1) The equation ((-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 a Sum or two is missing. $\endgroup$
    – Michael E2
    Commented Feb 24, 2022 at 16:21

1 Answer 1

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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.

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  • $\begingroup$ thanks I corrected the sum notation. The sum over 𝑖 goes from 0 to 𝑘. Trying to solve the sum using the induction steps also seems like a challenge to me. $\endgroup$
    – janhardo
    Commented Feb 23, 2022 at 15:23
  • $\begingroup$ Glad I was able to help. Hopefully the answer is clear! $\endgroup$
    – user49048
    Commented Feb 23, 2022 at 15:24
  • $\begingroup$ Have to study the answer first. I do see at the end of the answer a sequence of numbers, from which a general expression in k is derived. $\endgroup$
    – janhardo
    Commented Feb 23, 2022 at 15:29
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    $\begingroup$ @janhardo but this is a site for a Mathematica derivation. This is what I provided. For a derivation by hand, perhaps it's more appropriate to ask on the Mathematics site. $\endgroup$
    – user49048
    Commented Feb 23, 2022 at 16:35
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    $\begingroup$ @ DiSp0sablE_H3r0, i am complete satisfied with your answer , thanks for that!. Its "Complete induction" and the fun is trying to do it with MMA $\endgroup$
    – janhardo
    Commented Feb 23, 2022 at 17:17

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