Question

Let $S_n(1\le n\le 9)$ denote the sum of $n$ terms of the series $1+22+333+...+\frac{\mathrm{999...9}}{9times}$ then for $2\le n\le 9$

• A. $S_n-S_{n-1}=\frac{1}{9}({10}^n-n^2+n)$
• B. $S_n=\frac{1}{9}({10}^n-n^2+2n-2)$
• C. $9(S_n-S_{n-1})=n({10}^n-1)$
• D. none of these

Answer 1

The correct option is C $9(S_n-S_{n-1})=n({10}^n-1)$

Given $S_n=1+22+333+...+\mathrm{999...9}\left(9times\right)for(1\le n\le 9)$
For $n=9$,
$S_n=1+22+333+...+\mathrm{888...8}\left(8times\right)+\mathrm{999...9}\left(9times\right)$
Therefore, $S_{n-1}=1+22+333+...+\mathrm{888...8}\left(8times\right)$
Consider, ${(S}_n-S_{n-1})=(1+22+333+...+\mathrm{888...8}\left(8times\right)+\mathrm{999...9}\left(9times\right))-(1+22+333+...+\mathrm{888...8}\left(8times\right))$
$\Rightarrow {(S}_n-S_{n-1})=\mathrm{999...9}$
$\Rightarrow {(S}_n-S_{n-1})=9(\mathrm{111...1})$
$\Rightarrow {(S}_n-S_{n-1})=9({10}^8+{10}^7+{10}^6+...+{10}^0)$
$\Rightarrow {(S}_n-S_{n-1})=\frac{9({10}^n-1)}{9}$
$\Rightarrow 9{(S}_n-S_{n-1})=9({10}^n-1)$
$\Rightarrow 9{(S}_n-S_{n-1})=n({10}^n-1)....(sincen=9)$
Similarly, for $n=8$,
$S_n=1+22+333+...+\mathrm{888...8}\left(8times\right)$
$\Rightarrow S_n=1+22+333+...+\mathrm{777...7}\left(7times\right)+\mathrm{888...8}\left(8times\right)$
Therefore, $S_{n-1}=1+22+333+...+\mathrm{777...7}\left(7times\right)$
Consider, ${(S}_n-S_{n-1})=(1+22+333+...+\mathrm{777...7}\left(7times\right)+\mathrm{888...8}\left(8times\right))-(1+22+333+...+\mathrm{777...7}\left(7times\right))$
$\Rightarrow {(S}_n-S_{n-1})=\mathrm{888...8}$
$\Rightarrow {(S}_n-S_{n-1})=8(\mathrm{111...1})$
$\Rightarrow {(S}_n-S_{n-1})=8({10}^7+{10}^6+{10}^5+...+{10}^0)$
$\Rightarrow {(S}_n-S_{n-1})=\frac{8({10}^n-1)}{9}$
$\Rightarrow 9{(S}_n-S_{n-1})=8({10}^n-1)$
$\Rightarrow 9{(S}_n-S_{n-1})=n({10}^n-1)....(sincen=8)$
In general, $9{(S}_n-S_{n-1})=n({10}^n-1)$
Thus, $9{(S}_n-S_{n-1})=n({10}^n-1)$, $for(2\le n\le 9)$.