Question

Let $f\left(n\right)=\sum _{r=1}^n{r}^2{\left({}^n{C}_r\right)}^2$ then the value of $f\left(5\right)$ is equal to

• A. $1000$
• B. $1250$
• C. $1750$
• D. $2500$

Answer 1

The correct option is C $1750$

$f\left(n\right)={\sum }_{r=1}^n{r}^2{(}^n{C}_r{)}^2$ then f(5)

$\therefore f\left(5\right)={\sum }_{r=1}^5{r}^2{(}^5{C}_r)$

$=(1{)}^2{(}^5{C}_1{)}^2+(2{)}^2{(}^5{C}_2{)}^2+(3{)}^2{(}^5{C}_3{)}^2+(4{)}^2{(}^5{C}_4{)}^2+(5{)}^2{(}^5{C}_5{)}^2$

$=1\times (5{)}^2+4\times {\left(\frac{5\times 4}{2\times 1}\right)}^2+9\times {\left(\frac{5\times 4\times 3}{3\times 2\times 1}\right)}^2+16\times {\left(\frac{5\times 4\times 3\times 2}{4\times 3\times 2\times 1}\right)}^2+25\times (1{)}^2$

$=25+4\times 100+900+25\times 16+25$

$=50+400+900+400$

$=1750$