General solution for $u_{n+1}-u_{n}=7\cdot 3^n+5(2n^2-n)$?

Question

How to obtain a general solution for:

$$u_{n+1}-u_{n}=7\cdot 3^n+5(2n^2-n)\quad?$$

Answer 1

We look for a solution of the form $$u_n=A3^n+Bn^3+Cn^2+Dn+E$$ where $A$, $B$, $C$, $D$ and $E$ are constants. Then \begin{eqnarray*} u_{n+1} & = & A3^{n+1}+B(n+1)^3+C(n+1)^2+D(n+1)+E\\ & = & 3A3^n+Bn^3+(3B+C)n^2+(3B+2C+D)n+B+C+D+E \end{eqnarray*} so that $$u_{n+1}-u_n=2A3^n+3Bn^2+(3B+2C)n+(B+C+D)=7\cdot 3^n+10n^2-5n$$ Now you can solve the equations: $$\begin{array}{rcl} 2A & = & 7 \\ 3B & = & 10 \\ 3B+2C & = & -5 \\ B+C+D & = & 0 \end{array}$$ to find the constants.

Answer 2

Exploit telescopy.

$\sum_{k=0}^{k=n-1}(u_{k+1}-u_{k})=u_n-u_0=\sum_{k=0}^{k=n-1}(7\times3^k+5\times(2\times k^2-k))=7\times\frac{3^{n}-1}{2}+5\times(\frac 13 n(n-1)(2n-1)-\frac 12n(n-1))=\frac 72(3^n-1)+\frac 56 n(n-1)( 4n-5)$

Answer 3

Let us add the $n$ first equalities of the form given in the question (what is called "Telescopic cancellation"). We get:

$$\tag{a}u_{n+1}-u_1=7 \frac{3^{n+1}-1}{3-1}+10\frac{n(n+1)(2n+1)}{6}-5\frac{n(n+1)}{2}$$

providing an explicit expression of the general term (if we assume that the sequence begins at index 1).

Remark: the formulas used in the RHS of relationship (a) come

• from the formula of the sum of a finite geometric sequence and

• from (21) and (22) in (http://mathworld.wolfram.com/PowerSum.html).

Answer 4

We have:

$$u_{n+1}=7\times3^n+10n^2-5n+u_n$$

$$=\underbrace{\left[7\times3^n+10n^2-5n\right]}_n+\underbrace{\left[7\times3^{n-1}+10(n-1)^2-5(n-1)\right]}_{n-1}+u_{n-1}$$

Repeated expansion gives us

$$u_{n+1}=u_0+\sum_{k=0}^n\left[7\times3^k+10k^2-5k\right]$$

$$=u_0+\frac72\left(3^{n+1}-1\right)+10\frac{n(n+1)(2n+1)}6-5\frac{n(n+1)}2$$

$$u_{n+1}=u_0+\frac72\left(3^{n+1}-1\right)+\frac{n(n+1)(20n-5)}6$$