Problem Analysis #common_20

Recorded Python call:

# Verification of the solution
# Given recurrence: a_1 = 7
# S_n = sum_{k=1}^n a_k = (2/3)a_n + (1/6)n^2 - (1/6)n + 10 for n>=2
# We derived: a_{n+1} = (2/3)(a_{n+1} - a_n) + n/3  => (가) = n/3
# So f(n) = n/3, f(12) = 4.
# Compute a_2 using n=2: S_2 = a_1 + a_2 = 7 + a_2
# RHS: (2/3)a_2 + (1/6)*4 - (1/6)*2 + 10 = (2/3)a_2 + 4/6 - 2/6 + 10 = (2/3)a_2 + 1/3 + 10 = (2/3)a_2 + 31/3
# 7 + a_2 = (2/3)a_2 + 31/3 => 21 + 3a_2 = 2a_2 + 31 => a_2 = 10  => p=10
# Recurrence from ㉠: 2a_n + a_{n+1} = n for n>=2.
# Sum needed: S_12 + sum_{k=1}^5 a_{2k+1} = a_1 + a_2 + sum_{k=1}^5 (2a_{2k+1} + a_{2k+2})
# For each k, n=2k+1 >=3, so 2a_{2k+1} + a_{2k+2} = 2k+1.
# Sum_{k=1}^5 (2k+1) = 3+5+7+9+11 = 35.
# Total = 7 + 10 + 35 = 52  => q=52.
# Then (p*q)/f(12) = (10*52)/4 = 130.

# Let's also verify by computing all a_n up to 12 using recurrence a_{n+1} = n - 2a_n for n>=2.
a = [0]*13
a[1] = 7
a[2] = 10
for n in range(2, 12):
    a[n+1] = n - 2*a[n]

print("a_n:", a[1:])

# Compute S_12
S12 = sum(a[1:13])
print("S_12:", S12)

# Compute sum of a_{2k+1} for k=1..5
sum_odd = sum(a[2*k+1] for k in range(1, 6))
print("sum a_{2k+1}:", sum_odd)

total = S12 + sum_odd
print("Total:", total)

# Verify the expression a_1 + a_2 + sum(2a_{2k+1} + a_{2k+2})
alt_sum = a[1] + a[2] + sum(2*a[2*k+1] + a[2*k+2] for k in range(1, 6))
print("Alternative sum:", alt_sum)

# f(12)
f12 = 12/3
print("f(12):", f12)

# Result
p = 10
q = 52
result = (p * q) / f12
print("Result:", result)

The sequence is defined by $a_1=7$ and for $n\ge 2$,

Finding (가) and $f(n)$:
For $n\ge 2$,

Since $(n+1)^2-n^2=2n+1$ and $(n+1)-n=1$, the last two terms simplify to $\frac{1}{6}(2n+1)-\frac{1}{6}=\frac{n}{3}$.
Hence $a_{n+1}=\frac{2}{3}(a_{n+1}-a_n)+\frac{n}{3}$, so (가) $= \frac{n}{3}$. Thus $f(n)=\frac{n}{3}$ and $f(12)=\frac{12}{3}=4$.

Finding $p=$ (나):
Set $n=2$ in the sum formula:

With $a_1=7$, we get $7+a_2=\frac{2}{3}a_2+\frac{31}{3}\Rightarrow 21+3a_2=2a_2+31\Rightarrow a_2=10$. So $p=10$.

Finding $q=$ (다):
From the rearranged relation,

Now evaluate the required sum:

For $k=1,\dots,5$, set $n=2k+1\ge3$ in $2a_n+a_{n+1}=n$:

Thus

Hence the total equals $a_1+a_2+35=7+10+35=52$. So $q=52$.

Final computation: