How to solve the general formula by the recursive formula in MMA?

Question

I want to solve this problem $T(n)=\sum_{i=1}^{n-1} 2T(i),T(1)=1$, I want to know how to solve this in MMA?

My thinking:

I have tried RSolve command, but the output is the same as the input.

$$RSolve[a[n]==\sum_{i=1}^{n-1} 2*a[i],a,n]$$

Answer 1

It is easy to deal with this problem in the numerical way. Besides the approach involved with memoization as Bob Hanlon used, another is to use Nest:

Nest[Append[#, 2 Total[#]] &, {1}, 10]
FindSequenceFunction[%[[2 ;;]], n]

generate

{1, 2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366}

2 3^(-1 + n)

This means that $ T(n) = 2\times 3^{n-2} $ for $ n = 2,3,4,... $.

Answer 2

This can also be seen as $T(n+1)-T(n)=2 T(n)\iff T(n+1)=3 T(n)$ for $n>1$, so$T(n)= 3T(n-1)=9 T(n-2)=...=3^{n-2}T(2)$ and $T(2)=2$. Therefore, $ T(1)=1,T(n)=2\cdot 3^{n-2}, for \quad n>1$

Answer 3

Clear[T];
T[1] = 1;
T[n_Integer?Positive] := T[n] = Sum[2*T[i], {i, 1, n - 1}];

Generate a sequence from the recursion

seq = T /@ Range[10]

(* {1, 2, 6, 18, 54, 162, 486, 1458, 4374, 13122} *)

Use FindSequenceFunction to find the general formula

f[n_] = FindSequenceFunction[seq, n]

The general formula is expressed as a DifferenceRoot

Verifying that the recursion and the general formula are equivalent even for values outside the original sequence:

(T /@ Range[100]) == (f /@ Range[100])

(* True *)

EDIT: As pointed out by Αλέξανδρος Ζεγγ for n > 1

Clear[T2];
T2[1] = 1;
T2[n_Integer?Positive] = FindSequenceFunction[{#, T[#]} & /@ Range[2, 10], n]

(* 2 3^(-2 + n) *)

Checking,

(T /@ Range[100]) == (f /@ Range[100]) == (T2 /@ Range[100])

(* True *)

Answer 4

"T[n] == Sum[2 T[i], {i, 1, n - 1}]" isn't really a (single) recurrence equation, so RSolve won't do much with it. (It's a different recurrence equation for each choice of $n$.) It is analogous to asking DSolve to do something useful with $f(x) = \int_{1}^{x}f(u) \,\mathrm{d}u$, which it won't because there is no derivative here. Here, we ask DSolve to solve an integral equation (or generally, an integrodifferential equation) in the same way we have asked RSolve to solve a summatory equation (or generally, a summation-recurrence equation).

Other solutions here have suggested first manipulating the equation by hand, then using RSolve. This is to convert your equation into a (single) recurrence equation. To convert the analogous integral to something DSolve will make progress with, we make Mathematica apply the fundamental theorem of calculus for us, then DSolve makes progress.

D[#,x]& /@ (f[x] == Integrate[f[u],{u,1,x}])
DSolve[%, f[x], x]
(*  f'[x] = f[x]  
    {{ f[x] -> E^C[1] }}
 *)

We need to do the same thing here (and we'll include T[1]==1).

DifferenceDelta[#, n]& /@ (T[n] == Sum[2 T[i], {i, 1, n - 1}])
RSolve[{ %, T[1]==1}, T[n], n]
(*  -T[n] + T[1 + n] == 2 T[n]
    {{ T[n] -> 3^(-1+n) }}
 *)

Answer 5

Using @ubdqn's observations with RSolve:

RSolve[{t[n] == 3 t[n - 1], t[2] == 2}, t[n], n]

{{t[n] -> 2 3^(-2 + n)}}