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Question
Find the sum of the following series:
9 + 99 + 999 + ... to n terms;
Solution
We have,
9 + 99 + 999 + ... n terms
\[= \left( 9 + 99 + 999 + . . . + \text { to n terms } \right)\]
\[ = \left\{ \left( 10 - 1 \right) + \left( {10}^2 - 1 \right) + \left( {10}^3 - 1 \right) + . . . + \left( {10}^n - 1 \right) \right\}\]
\[ = \left\{ \left( 10 + {10}^2 + {10}^3 + . . . + {10}^n \right) \right\} - \left( 1 + 1 + 1 + 1 . . .\text { n times } \right)\]
\[ = \left\{ 10 \times \frac{\left( {10}^n - 1 \right)}{10 - 1} - n \right\} \]
\[ = \left\{ \frac{10}{9}\left( {10}^n - 1 \right) - n \right\}\]
\[ = \frac{1}{9}\left\{ {10}^{n + 1} - 9n - 10 \right\}\]
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