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Question

Find the sum of n terms of the series.
0.5+0.55+0.555+....n terms.

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Solution

Given series 0.5+0.55+0.555...n terms
we know that,
0.1+0.12+0.13+....=0.1(10.1n)0.9=(10.1n)95(0.1+0.11+0.111+..)5(110+11100+1111000+...)59(910+99100+9991000+..)59((1110)+(11100)+(111000)+..)59((1+1+...n terms)(110+1100+11000+...))59(n(10.1n)9)

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