Solve $T(n) =2T(n-5)+n$, $T(1)=1$ and proof the solution is true by induction I've seen the solution of this Solve the recurrence $T(n) = 2T(n-1) + n$
So I have done 
1) $T(n)=2T(n-5)+n$ , $T(n-5)=2T(n-2\cdot5)+(n-5)$ e.t.c , my final line is
$$T(n)=2^kT(n-5k)+\sum_{j=0}^{k-5} 2^j(n-j).$$ 
Any help? From link I can't figure out how the sum works on my case.
 A: My compulsion for generality
leads me to consider
$t(n) = at(n-m)+f(n)
$.
Then
$\begin{array}\\
t(n) 
&= at(n-m)+f(n)\\
&= a(at(n-2m)+f(n-m))+f(n)\\
&= a^2t(n-2m)+af(n-m)+f(n)\\
&= a^2(at(n-3m)+f(n-2m))+af(n-m)+f(n)\\
&= a^3t(n-3m)+a^2f(n-2m)+af(n-m)+f(n)\\
...\\
&= a^kt(n-km)+\sum_{j=0}^{k-1}a^jf(n-jm)\\
&\text{induction step:}\\
&= a^k(at(n-(k+1)m)+f(n-km))+\sum_{j=0}^{k-1}a^jf(n-jm)\\
&= a^{k+1}t(n-(k+1)m)+a^kf(n-km)+\sum_{j=0}^{k-1}a^jf(n-jm)\\
&= a^{k+1}t(n-(k+1)m)+\sum_{j=0}^{k}a^jf(n-jm)\\
\end{array}
$
Setting
$k 
=\lfloor \dfrac{n}{m} \rfloor
$
and
$h = n-km$,
we have
$0 \le h \le m-1
$.
Then
$t(n)
=a^kt(n-km)+\sum_{j=0}^{k-1}a^jf(n-jm)
=a^kt(h)+\sum_{j=0}^{k-1}a^jf(n-jm)
$.
If $f(n) = n$,
$\begin{array}\\
t(n)
&=a^kt(h)+\sum_{j=0}^{k-1}(n-jm)a^j\\
&=a^kt(h)+n\sum_{j=0}^{k-1}a^j-m\sum_{j=0}^{k-1}ja^j\\
&=a^kt(h)+n\dfrac{a^k-1}{a-1}-m\dfrac{(k - 1) a^{k + 1} - k a^k + a}{(a - 1)^2}\\
&=a^kt(h)+\dfrac{n(a-1)(a^k-1)-m((k - 1) a^{k + 1} - k a^k + a)}{(a - 1)^2}\\
&=a^kt(h)+\dfrac{n(a^{k+1}-a^k-a+1)-m((k - 1) a^{k + 1} - k a^k + a)}{(a - 1)^2}\\
&=a^kt(h)+\dfrac{a^{k+1}(n-m(k-1))+a^k(km-n)-n(a-1)-am}{(a - 1)^2}\\
&=a^kt(h)+\dfrac{a^{k+1}(h+m)-ha^k-n(a-1)-am}{(a - 1)^2}\\
\end{array}
$
A: Assume that $n=5a+b$ with $0\leq b<5$, with $T(n)=2T(n-5)+n$ for $a=\lfloor n/5\rfloor>0$, then
$$\boxed{\quad T(n)=2^a(T(b)+b+10)-n-10\quad}$$
We verify the solution by induction.
It works for $n=b$ (i. e. $a=0$), and, for the inductive step,
$$T(n+5)=2T(n)+n+5=2(2^a(T(b)+b+10)-n-10)+n+5
\\=2^{a+1}(T(b)+b+10)-(n+5)-10.$$
