The sum S can also be expressed as
[tex]S = \displaystyle \sum_{k=1}^n k\cdot5^{k-1}[/tex]
As a first step, pull out a factor of 5 from the sum:
[tex]S = \displaystyle 5 \sum_{k=1}^n k\cdot 5^{k-2}[/tex]
Shift the index to force the sum to start at k = 0, then distribute the summation:
[tex]S = \displaystyle 5 \sum_{k=0}^{n-1} (k+1) 5^{k-1} \\\\ S = 5\sum_{k=0}^{n-1}k\cdot5^{k-1} + 5\sum_{k=0}^{n-1}5^{k-1} \\\\ S = 5\sum_{k=0}^{n-1}k\cdot5^{k-1} + \sum_{k=0}^{n-1}5^k[/tex]
The second sum is geometric, with
[tex]\displaystyle \sum_{k=0}^{n-1} 5^k = 1 + 5 + 5^2 + \cdots + 5^{n-1} \\\\ \implies 5\sum_{k=0}^{n-1} 5^k = 5 + 5^2 + 5^3 + \cdots + 5^n \\\\ \implies \sum_{k=0}^{n-1} 5^k - 5\sum_{k=0}^{n-1} 5^k = 1 - 5^n \\\\ \implies \sum_{k=0}^{n-1} 5^k = \frac{5^n-1}4[/tex]
This leaves us with
[tex]\displaystyle S = 5\sum_{k=0}^{n-1}k\cdot5^{k-1} + \frac{5^n-1}4[/tex]
For the remaining sum, add and subtract the k = n-th term, so that we have
[tex]\displaystyle S = 5\left(\sum_{k=0}^n k\cdot 5^{k-1} - n\cdot5^{n-1}\right) + \frac{5^n-1}4[/tex]
Then in the sum, we get 0 for the k = 0 term and end up recovering another copy of S :
[tex]\displaystyle S = 5\left(S - n\cdot5^{n-1}\right) + \frac{5^n-1}4 \\\\ S = 5S - n\cdot5^n + \frac{5^n-1}4[/tex]
Solving for S gives
[tex]-4S = \dfrac{5^n-1}4 - n\cdot5^n \\\\ S = \dfrac{n\cdot5^n}4 - \dfrac{5^n-1}{16} \\\\ \boxed{S = \dfrac{(4n-1)5^n-1}{16}}[/tex]
