MHB A=1×3×5×7+3×5×7×9+5×7×9×11+−−−+(2n−3)×(2n−1)×(2n+1)×(2n+3)

[Albert1]

$A=1\times 3\times 5\times 7+3\times 5\times 7\times 9+5\times 7\times 9\times 11
+---+(2n-3)\times (2n-1)\times (2n+1)\times (2n+3)$
here $n\geq 2, and \,\, n\in N$
$simplify\,\,A\,\, in\,\, expression \,\, with \,\,n$

 

[MarkFL]

My solution:

Spoiler

We may express $A$ recursively as follows:

$$A_{n}-A_{n-1}=(2n-3)(2n-1)(2n+1)(2n+3)=16n^4-40n^2+9$$ where $2\le n$

Our homogeneous solution is:

$$h_n=k_1$$

And our particular solution will take the form:

$$p_n=An^5+Bn^4+Cn^3+Dn^2+En$$

Substituting into our recursion, there results:

$$5An^4-(10A-4B)n^3+(10A-6B+3C)n^2-(5A-4B+3C-2D)n+(A-B+C-D+E)=16n^4+0n^3-40n^2+0n+9$$

Equating corresponding coefficients, and solving the resulting system, we obtain:

$$(A,B,C,D,E)=\left(\frac{16}{5},8,-8,-20,\frac{9}{5}\right)$$

And so our particular solution is:

$$p_n=\frac{16}{5}n^5+8n^4-8n^3-20n^2+\frac{9}{5}n$$

Now, we know fromn the principle of superposition and the initial condition:

$$A_2=90+k=105\implies k=15$$

And so our solution (closed-form for $A_n$) is:

$$A_n=\frac{16}{5}n^5+8n^4-8n^3-20n^2+\frac{9}{5}n+15$$

 

[Albert1]

My solution:

Spoiler

We may express $A$ recursively as follows:

$$A_{n}-A_{n-1}=(2n-3)(2n-1)(2n+1)(2n+3)=16n^4-40n^2+9$$ where $2\le n$

Our homogeneous solution is:

$$h_n=k_1$$

And our particular solution will take the form:

$$p_n=An^5+Bn^4+Cn^3+Dn^2+En$$

Substituting into our recursion, there results:

$$5An^4-(10A-4B)n^3+(10A-6B+3C)n^2-(5A-4B+3C-2D)n+(A-B+C-D+E)=16n^4+0n^3-40n^2+0n+9$$

Equating corresponding coefficients, and solving the resulting system, we obtain:

$$(A,B,C,D,E)=\left(\frac{16}{5},8,-8,-20,\frac{9}{5}\right)$$

And so our particular solution is:

$$p_n=\frac{16}{5}n^5+8n^4-8n^3-20n^2+\frac{9}{5}n$$

Now, we know fromn the principle of superposition and the initial condition:

$$A_2=90+k=105\implies k=15$$

And so our solution (closed-form for $A_n$) is:

$$A_n=\frac{16}{5}n^5+8n^4-8n^3-20n^2+\frac{9}{5}n+15$$

nice job ,MarkFL !