Find the sum to n terms of the series 0.4+0.44+0.444+...to n terms
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Answer:
4(0.1+0.11+0.111+.....to n terms)
4/9{0.9+0.99+0.999....to n terms}
4/9{(1-0.1)+(1-0.01).(1-0.001)+... To n terms}
4/9{n-[1/10+1/100+1/1000+...to n terms]}
4/9{n - a(1-r^n)/1-r}
4/9{ n - 1/10[1 - (1/10)^n] ÷ 1 - 1/10}
4/9{n - 1/9(1-1/10^n)}
4n/9 - 4/81 [1-(1/10)^n]
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