The value of 1+(1+1/3)×1/4+(1+1/3+1/9)×1/16+(1+1/3+1/9+1/27)×1/64+⋯ is:

Question

UPSC Quizroom · Accepted Answer

Step 1: Recognize the pattern

Numerator of each term is a geometric series: 1 + 1/3 + 1/9 + …

Denominator: 1, 4, 16, 64, … = 4^{n−1}

So:
S = Σ_{n=1 to ∞} (Σ_{k=0 to n−1} (1/3)^k) × 1/4^{n−1}

Step 2: Sum the inner geometric series

Σ_{k=0 to n−1} (1/3)^k = (1 − (1/3)^n) / (1 − 1/3) = (3/2)(1 − (1/3)^n)

Step 3: Rewrite S

S = Σ_{n=1 to ∞} (3/2)(1 − (1/3)^n) × 1/4^{n−1}
= (3/2) × Σ_{n=1 to ∞} (1/4^{n−1} − 1 / (3^n × 4^{n−1}))

Step 4: Split into two geometric series

Σ 1/4^{n−1} = 1 + 1/4 + 1/16 + … = 1 / (1 − 1/4) = 4/3

Σ 1 / (3^n × 4^{n−1}) = Σ 4 / (12^n) = 4 × (1/12) / (1 − 1/12) = 4 / 11

Step 5: Compute final sum

S = (3/2) × (4/3 − 4/11) = (3/2) × (32/33) = 48 / 33 = 16 / 11

Answer: 16/11

🎯 UPSC Quiz Room