Problem Analysis #calculus_29

Recorded Python call:

from sympy import Rational, symbols, solve, simplify

# Interpretation: {a_n} is arithmetic with first term = common difference  =>  a_n = n*a_1, a_1 != 0
x = symbols('x')  # x := 1/a_1
lhs = (x/2 - 1)**2          # (1/a_2 - 1)^2
rhs = (x - 1)*(x/3 - 1)     # (1/a_1 - 1)(1/a_3 - 1)  (geometric: middle^2 = product of neighbors)
print("equation:", (lhs - rhs).expand(), "= 0")
roots = solve(lhs - rhs, x)
print("roots for x = 1/a1:", roots)

a1 = Rational(1, 4); d = a1
a = lambda n: n*a1
vals = [1/a(i) - 1 for i in (1, 2, 3)]
print("1/a_i - 1 :", vals)
r = simplify(vals[1]/vals[0])
print("ratio r =", r, " consistency b_{k+3}/b_{k+2} =", simplify(vals[2]/vals[1]))

# telescoping check: 1/(a_n a_{n+1}) = (1/d)(1/a_n - 1/a_{n+1})  => sum = 1/(d*a1) = 16
N = 200000
T = sum(1.0/(a(n)*a(n+1)) for n in range(1, N+1))
print(f"numeric partial sum T (N={N}) = {T:.6f}   exact 1/(d*a1) = {1/(d*a1)}")

print("k-scan:  b_{k+1}=3 = b1*r^k  =>  b1 = 3*3^k,  sum b_n = 9*3^k/2")
for k in range(1, 6):
    b1 = vals[0]*Rational(3)**k
    Sb = b1/(1 - r); diff = Sb - 16
    print(f"  k={k}: sum b_n = {Sb} = {float(Sb):.4f},  sum-16 = {float(diff):.4f},  0<diff<30 : {0 < diff < 30}")

k = 2
b1 = vals[0]*Rational(3)**k
b = lambda n: b1*r**(n-1)
print("k=2: b1 =", b1, "; check b_{k+i} == 1/a_i - 1 :", [b(k+i) for i in (1,2,3)], "vs", [1/a(i)-1 for i in (1,2,3)])

exact_even = simplify(b1*r/(1 - r**2))
partial_even = float(sum(b(2*n) for n in range(1, 60)))
print("sum_{n>=1} b_{2n}: exact =", exact_even, " partial =", partial_even)
res = simplify(a(2)*exact_even)
print("a_2 * sum b_{2n} =", res, "=", float(res), "  p =", res.p, " q =", res.q, "  p+q =", res.p + res.q)

Recorded Python call:

import sympy as sp
# Cross-check the alternative reading (b1=a1=c, arithmetic d = geometric ratio s): does it give any valid solution?
c, s = sp.symbols('c s', real=True)
for k in range(1, 7):
    eqs = [c*s**k - (1/c - 1),
           c*s**(k+1) - (1/(c+s) - 1),
           c*s**(k+2) - (1/(c+2*s) - 1)]
    sols = set()
    for c0 in [-5,-2,-1,-0.5,-0.2,0.2,0.5,1,1.5,2,5,10]:
        for s0 in [-0.95,-0.7,-0.4,-0.15,0.15,0.4,0.7,0.95]:
            try:
                sol = sp.nsolve(eqs, (c, s), (c0, s0), tol=1e-16, maxsteps=80, prec=30)
                cc, ss = float(sol[0]), float(sol[1])
                if abs(cc) < 1e-8 or abs(cc+ss) < 1e-8 or abs(cc+2*ss) < 1e-8:  # denominators
                    continue
                if abs(ss) < 1e-9 or abs(ss) >= 1 - 1e-9:  # need convergence of both series
                    continue
                resid = max(abs(float(e.subs({c: sol[0], s: sol[1]}))) for e in eqs)
                if resid < 1e-8:
                    sols.add((round(cc, 6), round(ss, 6)))
            except Exception:
                pass
    if sols:
        for (cc, ss) in sorted(sols):
            S = cc/(1-ss) - 1/(cc*ss)   # sum b_n - telescoped sum of 1/(a_n a_{n+1})
            print(f"alt k={k}: c={cc}, s={ss}, S={S:.6f}, 0<S<30: {0 < S < 30}")
    else:
        print(f"alt k={k}: no admissible solution")

해설 (검증 포함 요약)

• 첫째항 = 공차인 등차수열이므로 $a_n = n a_1$. 기하(등비) 조건 $(\tfrac1{a_2}-1)^2=(\tfrac1{a_1}-1)(\tfrac1{a_3}-1)$에서 $x=\tfrac1{a_1}$로 두면 $-\frac{x^2}{12}+\frac{x}{3}=0$, 즉 $x=4$ (0은 제외) → $a_1=\frac14,\ a_2=\frac12,\ a_3=\frac34$.
• 그러면 $b_{k+i}=\frac1{a_i}-1=(3,1,\tfrac13)$로 공비 $r=\frac13$인 등비수열이고, $b_1=3\cdot 3^k$.
• 망원합 검증(수치 부분합 15.99992 ≈ 16): $\sum \frac1{a_na_{n+1}}=\frac{1}{a_1^2}\cdot\frac{1}{1\cdot2}\cdot2=16$.
• 조건 $0<\sum b_n-16<30$을 만족하는 $k$는 스캔 결과 k=2뿐 ($\sum b_n=\frac{81}{2}$, 차이 $24.5$) → $b_1=27$.
• $\sum_{n\ge1}b_{2n}=\dfrac{b_1 r}{1-r^2}=\dfrac{27/3}{1-1/9}=\dfrac{81}{8}$, 따라서 $a_2\cdot\sum b_{2n}=\frac12\cdot\frac{81}{8}=\frac{81}{16}$ → $p+q=97$.

(대안 해석 — 공비가 공차와 같은 경우 — 에 대해서도 수치 다중 시작점 탐색을 수행했으나 수렴 가능한 해가 없어, 위 해가 유일함을 확인했습니다.)