Let be a sequence satisfying the following conditions.\begin{center}\fbox{\begin{minipage}{0.9\textwidth}
- ,
- For every natural number , \end{minipage}}\end{center}Consider the following work to compute\begin{center}\fbox{\begin{minipage}{0.9\textwidth}For every natural number ,soand, after rearranging this expression,Fromsubstituting into both sides givesBy (i) and (ii),\sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} = a_1 + a_2 + \sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2})$$$$= \boxed{(\text{C})}.\end{minipage}}\end{center}Let be the expression corresponding to \boxed{(\text{A})}, and let and be the numbers corresponding to \boxed{(\text{B})} and \boxed{(\text{C})}, respectively. Compute[4 points]\
[THOUGHT] Derive f(n) from the difference S_{n+1}-S_n, compute a2 from the n=2 case, use the derived relation 2a_n+a_{n+1}=n to evaluate the required sums (thus q), then compute (p*q)/f(12). I will verify all values with Python by symbolic simplification and by generating the sequence. [/THOUGHT] [PYTHON] import sympy as sp
n=sp.symbols('n', integer=True, positive=True) S = lambda m, a_m: sp.Rational(2,3)*a_m + sp.Rational(1,6)*m**2 - sp.Rational(1,6)*m + 10
derive A = polynomial difference part in a_{n+1} expression
A = sp.simplify((sp.Rational(1,6)(n+1)**2 - sp.Rational(1,6)(n+1) + 10) - (sp.Rational(1,6)*n**2 - sp.Rational(1,6)*n + 10)) print("A =", A)
compute a2 from n=2 equation
a2=sp.Symbol('a2') eq = sp.Eq(7 + a2, sp.Rational(2,3)*a2 + sp.Rational(1,6)*2**2 - sp.Rational(1,6)*2 + 10) p = sp.solve(eq, a2)[0] print("p (a2) =", p)
compute q using 2a_n+a_{n+1} = n and sum
q = 7 + p + sum([2*k+1 for k in range(1,6)]) print("q =", q)
f12 = sp.simplify(A.subs(n,12)) print("f(12) =", f12)
value = sp.simplify(pq/f12) print("pq/f(12) =", value)
extra verification by generating sequence using recurrence from relation:
from a_{n+1} = 2/3(a_{n+1}-a_n)+A => gives a_{n+1} = -2a_n + n
def gen_a(N): a=[None](N+1) a[1]=7 a[2]=int(p) for m in range(2,N): a[m+1] = -2a[m] + m return a
a=gen_a(12) print("a1..a12:", a[1:13])
verify given S_n formula for n>=2
def S_n(m): return sum(a[1:m+1]) for m in range(2,13): lhs=S_n(m) rhs=sp.Rational(2,3)*a[m] + sp.Rational(1,6)*m**2 - sp.Rational(1,6)m + 10 if sp.simplify(lhs-rhs)!=0: print("Mismatch at", m, lhs, rhs) print("Sum expression check:", sum(a[1:13]) + sum(a[2k+1] for k in range(1,6))) [/PYTHON][THOUGHT] The computations give (f(n)=\frac n3), (p=a_2=10), and (q=52). Thus (\dfrac{p q}{f(12)}=\dfrac{10\cdot 52}{4}=130). [/THOUGHT] \boxed{130}
[THOUGHT] From the Python output, (f(12)=4), (p=10), (q=52), so (\frac{p q}{f(12)}=\frac{10\cdot 52}{4}=130). \boxed{130}