The discussion focuses on the mathematical expression for A, defined as $A=1\times 3\times 5\times 7+3\times 5\times 7\times 9+5\times 7\times 9\times 11+---+(2n-3)\times (2n-1)\times (2n+1)\times (2n+3)$, where $n \geq 2$ and $n \in \mathbb{N}$. Participants aim to simplify this expression in terms of n. MarkFL received commendation for their solution, indicating a successful approach to the problem.
PREREQUISITESMathematicians, educators, and students interested in algebraic simplification and combinatorial mathematics will benefit from this discussion.
nice job ,MarkFL !MarkFL said:My solution:
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:
$$A_n=\frac{16}{5}n^5+8n^4-8n^3-20n^2+\frac{9}{5}n+15$$