Question

Find the $n^{th}$ term and the sum of first $n$ terms of the sequence $0.3$, $0.33$, $0.333$,$0.3333$,.....

Answer 1

$n^{th}$ term is $n$ number of $35$ after decimal point

sum of $n$ terms

$=0.3+0.33+..........n$ terms

$=3(0.1+0.11+\mathrm{0.111............}n$ terms)

$=\frac{3}{9}(0.9+0.59+........n$ terms)

$=\frac{3}{9}\left(\right(1-0.1)+(1-0.1)+(1-0.001)+.......$ n terms)

$=\frac{3}{9}(n-(0.1+\mathrm{0.01......}$ m terms))

$=\frac{3}{9}n-\frac{3}{9}\times 0.1\frac{(1-(0.1{)}^{th})}{1-0.1}$

$=\frac{3}{9}n-\frac{3}{81}\left({10.1}^n\right)=\frac{n}{3}-\frac{1}{27}(1-{0.1}^n)$

$=\frac{1}{27}(9n-1+(0.1{)}^n)$