Solve the recurrence $T(n) = 2T(n-1) + n, T(1)=1, n\geq 2$

Question

This question has been already solved here, I just want to figure out why I'm not being able to solve it using my method. Here's what I did -

$T(n)=2T(n-1)+n$
$T(n-1)=2T(n-2)+(n-1)$
$\therefore \,T(n)=2\{2T(n-2)+(n-1)\}+n$
$T(n)=4T(n-2)+3n-2$

$T(n-2)=2T(n-3)+(n-2)$
$\therefore T(n)=4\{2T(n-3)+(n-2)\}+3n-2$
$T(n)=8T(n-3)+7n-10$
$ \\. \\. \\. \\T(n)=2^kT(n-k)+(2^k-1)n-c \\\therefore T(n)=2^{n-1}T(1)+(2^{n-1}-1)n-c \\T(n)=2^{n-1}+(2^{n-1}-1)n-c $

This is where I'm stuck. I've seen other methods at the link above but I want to keep it simple by using induction. How should I proceed?
The answer is supposed to be : $2^{n+1}-n-2$

Why are you going this way? Are you interested in this technique? — Mhenni Benghorbal, Jan 30 '15 at 20:51
@MhenniBenghorbal, yes. I want to know why my method isn't producing the desired result. — Siddharth Thevaril, Jan 30 '15 at 20:53
You can't just say "oh, this is a constant, and I'll call it $c$" when you are approaching the problem with induction. Instead, you must guess a formula for $c$. — apnorton, Jan 30 '15 at 21:00
@anorton: Yes I'm aware of that, I was unable to guess a formula so I left it as $c$. Any help on how to? — Siddharth Thevaril, Jan 30 '15 at 21:01
It's similar to this; really, the first answer at the question you've linked gives the best approach. — apnorton, Jan 30 '15 at 21:08
@anorton: I know there are many ways to approach this but I would really like to know where my method went wrong. — Siddharth Thevaril, Jan 30 '15 at 21:13
@Sidsec9 After some more thought, the problem is that $c$ is not a constant, but is actually a function of $n$: $c(n) = n2^{n-1} + 2$. If you wrote out more terms, it is possible that you would have guessed the answer, but it's a bit complicated. You should be able to prove that with induction now. — apnorton, Jan 30 '15 at 21:18

abel · Accepted Answer · 2015-01-30T21:18:49.817

2

look for a particular solution of the form $T(n) = an + b$ where $a,b$ need to be determined. we need $an + b = 2(an - a + b) + n$ equating the coefficient of $n$ gives you $a = -2$ and then $b = 2b -a$ so $b = a + -2$ and $T(n) = -n-2$ is a particular solutions. the solution to the homogeneous equation is $T(n) = C2^{n}$ and the general solution is $$T(n) = C2^n - n - 2 $$ fix the $C$ to satisfy $T(1) = 1$

$\bf edit: $

we will need $1*2 + 2*2^2 + 3*2^3 + \cdots + (n-2)*2^{n-2} = (n-3)2^{n-1}+2$

here is your method taking from where you sopped

$\begin{align}T(n) &= 2^{n-1}T(1) + (2^{n-1} - 1)n -\left(1*2 + 2*2^2 + 3*2^3 + \cdots+ (n-2)2^{n-2} \right)\\ &=2^{n-1} + (2^{n-1} - 1)n -\left( (n-3)2^{n-1}+2 \right)\\ &=2^{n+1}-n -2 \end{align}$

edited Jan 30 '15 at 21:18

answered Jan 30 '15 at 20:56

abel

29,170

can you continue from where I left of? – Siddharth Thevaril Jan 30 '15 at 20:58
one question, did you stop at $(n-2)2^{n-2}$ because the constraint is $n\geq 2$ – Siddharth Thevaril Jan 30 '15 at 21:40
@Sidsec9, that is how many additions you will need to go from $T(n)$ to $T(1).$ – abel Jan 30 '15 at 21:52

Solve the recurrence $T(n) = 2T(n-1) + n, T(1)=1, n\geq 2$

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