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Find the Sum of the Following Series: 9 + 99 + 999 + ... to N Terms;

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प्रश्न

Find the sum of the following series:

9 + 99 + 999 + ... to n terms;

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उत्तर

 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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अध्याय 20: Geometric Progression - Exercise 20.3 [पृष्ठ २८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 20 Geometric Progression
Exercise 20.3 | Q 4.3 | पृष्ठ २८

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