Question

The value of $\frac{1}{270}{\sum }_{r=1}^8{r}^3{\left(\frac{{{}^{8}C}_r}{{{}^{8}C}_{r-1}}\right)}^2$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

$\frac{{}^n{c}_r}{{}^n{c}_{r-1}}=\left[\frac{n-r+1}{r}\right]$

$=1{\left[\frac{8}{1}\right]}^2+(2{)}^3{\left[\frac{7}{2}\right]}^2+(3{)}^3{\left[\frac{6}{3}\right]}^2+(4{)}^3{\left[\frac{5}{4}\right]}^2+(5{)}^3{\left[\frac{4}{5}\right]}^2+(6{)}^3{\left[\frac{3}{6}\right]}^2+(7{)}^3{\left[\frac{2}{7}\right]}^2+(8{)}^3{\left[\frac{1}{8}\right]}^2$

$=8^2+7^2\times 2+6^2\times 3+5^2\times 4+4^2\times 5+3^2\times 6+2^2\times 7+1\times 8$

$=540$

Value $=\frac{540}{270}=2$