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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) |
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Answer» (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. |
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