Sum of n terms of the series 1 + 11 + 111 + ... is

Question

Sum of n terms of the series 1 + 11 + 111 + ... is (A) (10/81)(10n-1)-(n/81) (B) (10/81)(10n-1)-(n/9) (C) (10/81)(10n-1)+(n/9) (D) (10/81)(10n-1)+(n/81)

Tardigrade Admin · Accepted Answer

Sn = 1+11+111+.....n terms = (1/9)[9+99+999...n terms] = (1/9) [(10-1)+(102-1)+(103-1)+...) = (1/9)[10 ((10n-1)/10-1)-n] = (10/81)[10n-1]-(n/9)

Q. Sum of n terms of the series $1 + 11 + 111 + ...$ is

$\frac{10}{81} (1 0^{n} - 1) - \frac{n}{81}$

$\frac{10}{81} (1 0^{n} - 1) - \frac{n}{9}$

$\frac{10}{81} (1 0^{n} - 1) + \frac{n}{9}$

$\frac{10}{81} (1 0^{n} - 1) + \frac{n}{81}$

$S_{n} = 1 + 11 + 111 + ..... n$ terms
$= \frac{1}{9} [9 + 99 + 999... n$ terms]
$= \frac{1}{9} [(10 - 1) + (1 0^{2} - 1) + (1 0^{3} - 1) + ...$ )
$= \frac{1}{9} [10 \frac{( 1 0 ^{n} - 1 )}{10 - 1} - n]$
$= \frac{10}{81} [1 0^{n} - 1] - \frac{n}{9}$

Q. Sum of n terms of the series 1+11+111+... is

8110​(10n−1)−81n​

8110​(10n−1)−9n​

8110​(10n−1)+9n​

8110​(10n−1)+81n​