Q:

Recurrence equation solution using reverse substitution method

Given
$$y(0)=7$$
$$y(n)=2y(n-1)+7^14^n$$
My attempt so far:
$$y(n)=2y(n-1)+7^14^n$$
$$=2^2y(n-2)+2^17^14^{n-1}+7^14^n$$
$$=2^3y(n-3)+2^27^14^{n-2}+2^17^14^{n-1}+7^14^n$$
$$=2^3y(n-3)+2^27^14^{n-2}+2^17^14^{n-1}+2^07^14^n$$
$$=2^iy(i)+2^{i-1}7^14^{n-i}+\dots$$
$$=2^ny(0)+2^{n-1}7^14^{n-n}+\dots$$
$$=7*2^n+n(2^{n-1}7)$$
However, if I calculate y(1) using the given y(n), I get 42 while I get 84 using the general term I found.
Wolframalpha suggest that the answer is 
$$y(n) = 2^{n - 1} (c_1 + 24 (2^n - 1))$$
Which step of my attempt is wrong ?

A:

You seem to have taken the last "dots" expression to be
$$2^{n-1}7^14^{n-n}+2^{n-1}7^14^{n-n}+\cdots+2^{n-1}7^14^{n-n}\ ,$$
that is, the same term $n$ times.  But it is actually
$$2^{n-1}7^14^{n-n}+2^{n-2}7^14^{n-(n-1)}+2^{n-3}7^14^{n-(n-2)}+\cdots\ ,$$
which can be added up using the formula for the sum of a geometric progression.
In fact there is also an error in the previous line, $2^{i-1}7^14^{n-i}$ should be $2^{i-1}7^14^{n-(i-1)}$.  If you make this correction and add up the new GP then you should get the right answer.

