```
result = Sum{k=1..n} (n+1-i) * k^3
= (n+1) * Sum{k=1..n} k^3 - Sum{k=1..n} k^4
```

(Sorry, no idea how to use mathematica symbols in HTML)

There are closed solutions for the two sums of k^a for any positive integral number a. You can find the solution for Sum(k^3) and a simple algorithm to derive the solution for Sum(k^4) at Sum of n, n², or n³ | Brilliant Math & Science Wiki[^]

The formula for k^3 as offered right at the top of that web site is:

`Sum{k=1..n} (k^3) = n^2*(n+1)^2 / 4`

and the sum for k^4 is given at the very bottom as:

`Sum{k=1..n} (k^4) = n*(n+1)*(2*n+1)*(3*n^2+3*n-1) / 30`

Now all you have to do is calculate both, multiply the former by (n+1). and subtract the latter.

P.S.: Here's a little program to prove the formulas are working as intended:

```
#include <iostream>
using namespace std;
double sum_of_power3(int n)
{
double dn = (double)n;
return dn*dn*(dn+1)*(dn+1)/4;
}
double sum_of_power4(int n)
{
double dn = (double)n;
return dn*(dn+1)*(2*dn+1)*(3*dn*dn+3*dn-1)/30;
}
int main()
{
double sn3 = 0;
double sn4 = 0;
for (int n = 1; n < 5; ++n)
{
sn3 += n*n*n;
sn4 += n*n*n*n;
cout << "sum of " << n << "^3 " << sn3 << " " << sum_of_power3(n) << endl;
cout << "sum of " << n << "^4 " << sn4 << " " << sum_of_power4(n) << endl;
}
return 0;
}
```

Output:sum of 1^3 1 1 sum of 1^4 1 1 sum of 2^3 9 9 sum of 2^4 17 17 sum of 3^3 36 36 sum of 3^4 98 98 sum of 4^3 100 100 sum of 4^4 354 354

Can you give reference to complete problem?