The sum of the first n terms in the sequence 8, 88, 888, 8888, ... is
(A) 80/81 (10^n - 1) + 8/9 n
(B) 80/81 (10^n - 1) - n
(C) 80/81 (10^n - 1) + 8n
(D) 80/81 (10^n - 1) - 8/9 n
Options:
(A) 80/81 (10^n - 1) + 8/9 n
(B) 80/81 (10^n - 1) - n
(C) 80/81 (10^n - 1) + 8n
(D) 80/81 (10^n - 1) - 8/9 n
Official Answer Key:
D
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2020 Q.8 / GA Q.8)**
**1. Algebraic Series Formulation**:
- Given series:
$S_n = 8 + 88 + 888 + \dots + \underbrace{88\dots 8}_{n \text{ digits}}$
- Factoring out $8$:
$S_n = 8 (1 + 11 + 111 + \dots)$
- Multiplying and dividing by $9$:
$S_n = \frac{8}{9} (9 + 99 + 999 + \dots) = \frac{8}{9} \sum_{k=1}^{n} (10^k - 1)$
**2. Detailed Summation**:
- Summing the geometric series $\sum_{k=1}^n 10^k$:
$\sum_{k=1}^n 10^k = \frac{10(10^n - 1)}{10 - 1} = \frac{10}{9}(10^n - 1)$
- Summing the constant term:
$\sum_{k=1}^n 1 = n$
- Substituting back into $S_n$:
$S_n = \frac{8}{9} \left[ \frac{10}{9}(10^n - 1) - n \right] = \frac{80}{81}(10^n - 1) - \frac{8}{9}n$
**3. Official Verification**:
The verified official IIT Delhi Master Answer is **Option (D)**.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.