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

Question

Solve the recurrence $T(n) = 2T(n-1) + n$ where $T(1) = 1$ and $n\ge 2$.

The final answer is $2^{n+1}-n-2$

Can anyone arrive at the solution?

Answer 1

\begin{align} T(n) & = 2 T(n-1) + n = 2(2T(n-2) + n-1) + n = 4T(n-2) + 2(n-1) + n\\ & = 8 T(n-3) + 4(n-2) + 2(n-1) + n = 2^k T(n-k) + \sum_{j=0}^{k-1} 2^j (n-j)\\ & = 2^{n-1} T(1) + \sum_{j=0}^{n-2}2^j (n-j) = 2^{n-1} + \sum_{j=0}^{n-2}2^j (n-j) \end{align} \begin{align} \sum_{j=0}^{n-2}2^j (n-j) & = n \sum_{j=0}^{n-2}2^j - \sum_{j=0}^{n-2} j2^j = n(2^{n-1}-1) - \dfrac{n \cdot 2^n - 3 \cdot 2^n + 4}2\\ & = n(2^{n-1}-1) - (n \cdot 2^{n-1} -3 \cdot 2^{n-1} + 2) = 3 \cdot 2^{n-1} -n - 2 \end{align} Hence, $$T(n) = 2^{n-1} + 3 \cdot 2^{n-1} -n - 2 = 2^{n+1} - n - 2$$

EDIT (Adding details)

First note that $\displaystyle \sum_{j=0}^{n-2}2^j$ is sum of a geometric progression and can be summed as shown below.$$\sum_{j=0}^{k} x^j = \dfrac{x^{k+1} -1}{x-1}$$ $\displaystyle \sum_{j=0}^{n-2} j2^j$ is a sum of the form $\displaystyle \sum_{j=0}^{k} jx^j$ $$\sum_{j=0}^{k} jx^j = x \sum_{j=0}^{k} jx^{j-1} = x \dfrac{d}{dx} \left( \sum_{j=0}^k x^j\right) = x \dfrac{d}{dx} \left( \dfrac{x^{k+1} - 1}{x-1}\right) = x \left( \dfrac{kx^{k+1} - (k+1) x^k +1}{(x-1)^2} \right)$$

Answer 2

Hint: substitute $T(n)=G(n)-n-2$

Answer 3

Induction: For $n=1$, $T(1)=1=2^{1+1}-1-2$. Suppose $T(n-1)=2^n-n+1-2=2^n-n-1$. Then $T(n)=2T(n-1)+n=2^{n+1}-2n-2+n=2^{n+1}-n-2$ which completes the proof.

Answer 4

Let's use the method of annihilators to turn this into a third-order, homogeneous recurrence, and then solve with a characteristic equation. First, we write the recurrence so $n$ is the least index: $$T(n) - 2T(n-1)= n \implies T(n+1)-2T(n) = n+1$$

Then, we rewrite the recurrence in terms of the shift operator $E$: $$(E-2)T(n) = n+1$$

Applying the $(E-1)$ operator to both sides of the equation, we have: $$(E-1)^2(E-2)T(n) = (E-1)^2(n+1)$$

Now, since $(E-1)^2(n+1) = 0$ (it annihilates that term), we have: $$(E-1)^2(E-2)T(n) = 0$$

The characteristic equation is, then $(r-1)^2(r-2)=0$, so our roots are $r=1$ (with multiplicity $2$) and $r=2$. So, the recurrence has the form: \begin{align*} T(n) &= c_1\cdot 2^n + c_2 \cdot n1^n + c_3 \cdot 1^n\\ &= c_1\cdot 2^n + nc_2 + c_3 \end{align*}

Using the recurrence relation, we compute $T(1) = 1, T(2) = 4, T(3) = 11$ and can now solve for $(c_1, c_2, c_3)$, which gives the solution $(2, -1, -2)$.

Answer 5

This recurrence $$T(n) = 2T(n-1) + n$$ is difficult because it contains $n$. Let $D(n) = T(n) - T(n-1)$ and compute $D(n+1) = 2D(n) + 1$ this recurrence is not so difficult. Of course $D(1) = 4 - 1 = 2 + 1$.

The sequence $D(n)$ goes: $2 + 1$, $2^2 + 2 + 1$, $2^3 + 2^2 + 2 + 1$. $D(n) = 2^{n+1}-1$.

Now $$T(n) = \sum_{i=1}^n D(i) = 2 \sum_{i=1}^n 2^{i} - \sum_{i=1}^n 1 = 2^{n+1}-2-n.$$

Answer 6

Some of the solutions presented here are very arcane. Building on the hint given be @Norbert, we consider the generalized problem

$${{T}_{n}}=A{{T}_{n-1}}+Bn,\quad{{T}_{1}}\,\,\,\text{specified}$$

Let $ {{T}_{n}}={{f}_{n}}+pn+q$ so that

$${{f}_{n}}=A{{f}_{n-1}}+n\left[Ap+B-p \right]+\left[ -Ap+Aq-q \right] $$

Then choose $p$ and $q$ so as to eliminate the quantities in brackets, i.e.,

$$p=-\frac{B}{A-1};\quad q=-\frac{AB}{{{\left( A-1 \right)}^{2}}} \\ {{f}_{n}}=A{{f}_{n-1}} \\ $$

Now, in order for $p$ and $q$, and hence the sequence to be integers, it must be that $A=2$. Clearly, the solution for $f$ is $f_n=2^{n-1}f_1$ and the general solution is given by

$$ \begin{align} T &=(T_1-p-q)2^{n-1}+pn+q\\ &=(T_1+3B)2^{n-1}-B(n+2) \end{align} $$

This result is in agreement with the others presented herein and has been tested numerically against the recurrence formula for arbitrary positive and negative integer values of $T_1$ and $B$.

Answer 7

By subtraction $$ \begin{array}{rrcl} & T(n)-2T(n-1) &=& n \\ & T(n-1) - 2T(n-2) &=& n-1 \\ \hline & T(n)-3T(n-1)+2T(n-2) &=& 1 \end{array}$$

Again, by subtraction

$$ \begin{array}{rrcl} & T(n)-3T(n-1)+2T(n-2) &=& 1 \\ & T(n-1)-3T(n-2)+ 2T(n-3) &=& 1 \\ \hline & T(n)-4T(n-1)+5T(n-2)-2T(n-3) &=& 0 \end{array}$$

Characteristic equation of the recursion is

$$x^3-4x^2+5x-2 = 0$$

Roots of the equation are $x_1 = x_2 = 1$ and $x_3 = 2$

So, general solution of the recursion is $$T(n) = C_1\cdot 1^n + C_2\cdot n\cdot 1^n + C_3\cdot 2^n$$ or $$T(n) = C_1 + C_2\cdot n + C_3\cdot 2^n$$

now, from $T(1) = 1$ we get $T(2) = 4$ and $T(3) = 11$.

So, we get system

$$ \begin{array}{ccccccc} C_1 &+&C_2 &+& 2C_3 &=& 1 \\ C_1 &+&2C_2&+&4C_3 & =& 4 \\ C_1 & +& 3C_2 &+& 8C_3 &=& 11 \end{array}$$

From this system we get that $C_1 = -2$, $C_2 =-1$, and $C_3 = 2$, so particular solution of the recursion is $$T(n) =2\cdot2^n-n-2,$$ or $$T(n) = 2^{n+1}-n-2.$$

Answer 8

T(n)=2T(n-1)+n -----(1)
T(n-1)=2T(n-2)+n-1 ------(2)
T(n-2)=2T(n-3)+n-2 ------(3)

Substitute (3) in (2) and then (2) in (1)

T(n-1)=$2^2T$(n-3)+2(n-2)+(n-1)

T(n)= $2^3$T(N-3)+$2^2$(n-2)+2(n-1)+n

=> T(k)=$2^k$T(n-k) + $2^{k-1}$(n-(k-1))+........+$2^0$n

Now, since T(1)=1, let n-k=1 => k=n-1

T(n)=$2^{n-1}$T(1) + $2^{n-2}$(2) + $2^{n-3}$(3) + $2^{n-n}$(n) --------(4)

Multiply Equation (4) with 2

T(n)=$2^{n}$+$2^{n-1}$ + $2^{n-2}$ + $2^{n-3}$+...... + 2n ----------- (5)

Subtract Equation (5) with (4)

T(n)=$2^{n}$+$2^{n-1}$+$2^{n-2}$+$2^{n-3}$+........+2 -n

Using sum of GP terms, we get :

[ $2^{1}$($2^{n}$-1) - n ]/ (2-1)

= $2^{n+1}$ -2 -n

Answer 9

Here's a generating-function approach to both derive and prove the desired formula. Write $T(n)$ as $T_n$, and let $f(z) = \sum_{n=1}^\infty T_n z^n$ be the ordinary generating function. Now use the recurrence relation and initial condition to obtain $$T_1 z^1 + \sum_{n=2}^\infty (T_n-2T_{n-1}) z^n = z + \sum_{n=2}^\infty n z^n,$$ equivalently, $$f(z)-2z f(z)=z + \left(z\sum_{n=1}^\infty n z^{n-1} - z\right) = z D_z \left(\sum_{n=1}^\infty z^n\right)=z D_z\left(\frac{1}{1-z}\right)=\frac{z}{(1-z)^2}.$$ Solving for $f(z)$ yields $$f(z) = \frac{z}{(1-2z)(1-z)^2}.$$ Now use partial fraction decomposition to obtain $$ f(z) = \frac{2}{1-2z}-\frac{1}{1-z}-\frac{1}{(1-z)^2} = \sum_{n=0}^\infty \left(2 \cdot 2^n - 1 - (n+1)\right)z^n, $$ which implies that $$T_n = 2 \cdot 2^n - 1 - (n+1) = 2^{n+1} - n - 2.$$

Answer 10

I expanded on "Siddharth Chakravorty" answer, not that his answer needed expanding. I could not understand the result of subtracting 2T(n) from T(n). After a long time, I got it. I shared my understanding of the result. Enjoy!

$Problem:$

$T(1) = 1$

$T(n) = 2T(n-1) + n$

$Equation \, 1 :$

$T(n) = 2T(n-1) + n$

$Equation \, 2$

$T(n-1) = 2T(n -1 -1) + n - 1$

$T(n-1) = 2T(n-2) + n - 1$

$Equation \, 3$

$T(n-2) = 2T(n-1-2) + n - 2$

$T(n-2) = 2T(n-3) + n -2$

$Back \, substitution$

$T(n) = 2T(n-1) + n$

$Substituted \, equation \, 2$

$T(n) = 2[2T(n-2) + n -1 ] + n$

$T(n) = 2^2 T(n-2) + 2(n-1) + n$

$Substituted \, equation \, 3$

$T(n) = 2^2 [2T(n-3) + (n - 2) ] + 2n -1 + n$

$T(n) = 2^3 T(n-2) + 2^2 (n -2) + 2(n -1) + n$

$ Continue for K \, times$

$T(n) = 2^k * T(n-k) + 2 ^{k-1} * (n-(k-1)) + ... + 2^0 n$

$Continue \, subtracting \; n -k \; until \, reaching \, 1$

$Let \, n - k = 1$

$-k = 1- n$

$k = -1 + n$

$k = n - 1$

$Substituting \; the \; value \; of \; k$

$T(n) = 2^{n-1} * T(n-(n -1)) + 2^{n-1-1} * (n - (n -1-1) + ... + 2^{n-n} * n$

$T(n) = 2^{n-1} * T(1) + 2^{n-2} * (n - (n -2)) + ... + 2^{0} * n$

$T(1) = 1$

$T(n) = 2^{n-1}(1) + 2^{n-2}(2) + ... + (n)$

$2 * T(n) = 2^1 * 2^{n-1}(1) + 2^1 * 2 ^{n-2} ( 2) + .. + 2^1(n)$

$2 * T(n) = 2^n(1) + 2^{n-1} (2)+ 2 ^{n-2} (3) + ... + 2^1(n)$

$Subtracting \; 2T(n) - T(n)$

$2 * T(n) - T(n)$

$Expanded \; T(n) \; to \, help \, with \, the \, subtraction$

$2^n + 2^{n-1} (2) + 2^{n-2} (3) + .... + 2^1(n) - [2^{n-1}(1) + 2^{n-2} (2) + 2^{n-3} (3) + .. + n]$

$2^n + 2^{n-1} (2-1) + 2^{n-2} (3-2) + 2^{n-3} (4 -3) + ...+ (- n)$

$2^n + 2^{n-1} + 2^{n-2} + 2^{n-3} + 2^{n-4} + ... + 2^0 - n$

$Geometric \, Progression \, Series$

$\frac{a (r^n - 1)}{(r - 1)}$

$a = 2$

$r = 2$

$\frac{2(2^n - 1)}{2 - 1} - n$

$2(2^n - 1) - n$

$2^{n + 1} - 2 - n$