Alg8.16

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Find the sum of the following sequence to n\, terms.

9,99,999,........(n9's)\,.

This is not a quite a GP.However, we can relate it to a GP by writing the terms as (10-1),(10^2-1),(10^3-1).........(10^n-1)\,

S_n=(10+10^2+10^3+...........)-(1+1+1+.......+n)\,

S_n=\frac{10(10^n-1)}{10-1}-n\,

Therefore S_n=\frac{10}{9}(10^n-1)-n\,


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