Question

Find the sum to 90 terms of the series $5+55+555+\cdots \cdots$

• A. $\frac{50}{81}[{10}^{90}-82]$
• B. $\frac{50}{81}[{10}^{90}-83]$
• C. $\frac{50}{81}[{10}^{90}-80]$
• D. $\frac{50}{81}[{10}^{90}-90]$

Answer 1

The correct option is A $\frac{50}{81}[{10}^{90}-82]$

Given

$5+55+555+5555+........+\underset{nfives}{\underset{}{\mathrm{55......5}}}$

$=\frac{5}{9}(9+99+999+9999+.......+\underset{nnines}{\underset{}{\mathrm{99......9}}}$

$=\frac{5}{9}({10}^1-1+{10}^2-1+{10}^3-1+.......+{10}^n-1)$

$=\frac{5}{9}({10}^1+{10}^2+{10}^3+......+{10}^n-n)$

$=\frac{5}{9}(\frac{{10}^{n+1}-10}{9}-n)$

Asked for $90$ terms

$n=90$

$=\frac{5}{9}(\frac{{10}^{91}-10}{9}-90)$

$=\frac{5}{9}(\frac{10({10}^{90}-1)}{9}-90)$

$=\frac{50}{9}(\frac{{10}^{90}-1}{9}-9)$

$=\frac{50}{9}\left(\frac{{10}^{90}-1-81}{9}\right)$

$=\frac{50}{81}({10}^{90}-82)$.