Sum the series :
1.1 + 3.01 + 5.001 + 7.0001 ..... n terms.

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Solution

5.001 = 5. $\frac{1}{10^{3}}$
The given series is A.G.P. $\frac{1}{10}, \frac{1}{10^{2}}, \frac{1}{10^{3}}, \dots i s \frac{1}{10} \cdot$
Ans. $n^{2} + \frac{1}{9} (1 - \frac{1}{10^{n}}) .$

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