33) Find the sum to n terms of the series 3 + 33 + 333 + ... to n terms.
Answer:
Let \(S_n = 3 + 33 + 333 + \dots\) to n terms.
Take 3 as a common factor:
\(S_n = 3(1 + 11 + 111 + \dots)\)
Multiply and divide by 9:
\(S_n = \frac{3}{9}(9 + 99 + 999 + \dots)\)
\(S_n = \frac{1}{3}[(10-1) + (10^2-1) + (10^3-1) + \dots \text{to n terms}]\)
Separate the terms into two series:
\(S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots \text{to n terms}) - (1 + 1 + 1 + \dots \text{to n terms})]\)
The first series is a Geometric Progression (G.P.) with first term \(a=10\) and common ratio \(r=10\). The sum is \(S_{GP} = \frac{a(r^n - 1)}{r-1}\).
\(S_n = \frac{1}{3} \left[ \frac{10(10^n - 1)}{10-1} - n \right]\)
\(S_n = \frac{1}{3} \left[ \frac{10}{9}(10^n - 1) - n \right]\)
Therefore, the sum is:
\[ S_n = \frac{10}{27}(10^n - 1) - \frac{n}{3} \]