Question

$S=\sum _{r=0}^n(-1{)}^r{}^n{C}_r[\frac{2^r}{3^r}+\frac{8^r}{3^{2r}}+\frac{{26}^r}{3^{3r}}+...\infty ]$ is

• A. $\frac{3^n}{3^n-1}$
• B. $1$
• C. $\frac{1}{3^n-1}$
• D. $\frac{2}{3^n-1}$

Answer 1

The correct option is C $\frac{1}{3^n-1}$

$S=\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{2}{3}\right)}^r+\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{8}{9}\right)}^r+\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{26}{27}\right)}^r+...\infty$

$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{2}{3}\right)}^r={(1-\frac{2}{3})}^n={\left(\frac{1}{3}\right)}^n=\frac{1}{3^n}$
$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{8}{9}\right)}^r={(1-\frac{8}{9})}^n={\left(\frac{1}{9}\right)}^n=\frac{1}{3^{2n}}$
$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{26}{27}\right)}^r={(1-\frac{26}{27})}^n={\left(\frac{1}{27}\right)}^n=\frac{1}{3^{3n}}$
$\therefore S=\frac{1}{3^n}+\frac{1}{3^{2n}}+\frac{1}{3^{3n}}+...\infty$
This is a geometric progression where
$a=\frac{1}{3^n},r=\frac{1}{3^n}$
$S=\frac{\frac{1}{3^n}}{1-\frac{1}{3^n}}=\frac{1}{3^n-1}$