The units digit of $(1 + 9 + 9^{2} + 9^{3} + . . . . + 9^{2009})$ is

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Solution

The correct option is A 0
here $9^{0} + 9^{1} = 10$ i.e units place is zero
again $9^{2} + 9^{3} = 810$ i.e units place is zero
$∴$ proceiding with the similar manner $9^{e v e n t e r m} + 9^{o d d t e r m}$ , the units place is zero
Hence, addition of pairs of all such terms will give units place 0.

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