Question 1144590
<pre>
{{{ S(n) = 1*5^1 + 2*5^2 }}} + ... + {{{ n*5^n }}}

Prove using induction, that the above is equal to
{{{ (5/16)*(4n*5^n - 5^n + 1) }}}

Base case:
S(1) = 5

and 

{{{ (5/16)*(4*1*5^1 - 5^1 + 1) = (5/16)*(20 - 5 + 1) = 5 }}}


Hypothesis:
Assume 
{{{ S(k) = (5/16)*(4k*5^k - 5^k + 1) }}}   (*)
holds for n=k.


Now let n=k+1:
{{{ S(k+1) = S(k) + (k+1)*5^(k+1) }}}

Re-writing the S(k) term using (*):
{{{ S(k+1) = ((5/16)*(4k*5^k - 5^k + 1)) }}} + {{{ (k+1)*5^(k+1) }}}

Now re-arrange (step-by-step shown):
{{{ S(k+1) = (5/16)*(4k*5^k - 5^k + 1)  + (k+1)*5^(k+1) }}}
{{{ S(k+1) = (5/16)*(4k*5^k - 5^k + 1  + (16/5)*(k+1)*5^(k+1)) }}}
{{{ S(k+1) = (5/16)*(4k*5^k - 5^k + 1  + (16k + 16)*5^k ) }}}
{{{ S(k+1) = (5/16)*(20k*5^k + 15*5^k + 1) }}}
{{{ S(k+1) = (5/16)*(4k*5^(k+1) + 3*5^(k+1) + 1) }}}

add and subtract {{{ 4*5^(k+1) }}}
{{{ S(k+1) = (5/16)*(4(k+1)*5^(k+1) - 4*5^(k+1) + 3*5^(k+1) + 1) }}}

finally:
{{{ S(k+1) = (5/16)*(4(k+1)*5^(k+1) - 5^(k+1) + 1) }}}

Thus (*) is true for n=k+1