Question

What is the sum of the 100 terms of the series 9 + 99 + 999 + ....?

(a) \(\frac{10}{9}\) (10¹⁰⁰ – 1) – 100 

(b) \(\frac{10}{9}\) (10⁹⁹ – 1) – 100 

(c) 100 (100¹⁰ – 1) 

(d) \(\frac{9}{100}\) (10¹⁰⁰ – 1)

Answer 1

(a) \(\frac{10}{9}\) (10¹⁰⁰ – 1) – 100 

Let S₁₀₀ = 9 + 99 + 999 + ...... upto 100 terms 

= (10 – 1) + (100 – 1) + (1000 – 1) + ..... + upto 100 terms 

= (10 + 10² + 10³ + .... upto 100 terms) – (1 + 1 + 1 + ..... upto 100 terms)

= \(\frac{10(10^{100}-1)}{10-1}-100\)                  \(\bigg(\because\,S_n=\frac{a(r^n-1)}{r-1}\,\text{when}\,r>1\bigg)\)

= \(\frac{10}{9}\) (10¹⁰⁰ – 1) – 100.