If  ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+\mathrm{.....}+C_n{x}^n,$

$\sum _{r=0}^n\left({(r+1)}^2\right)C_r=2^{n-2}f\left(n\right)$

and if the roots of the equation f(x) = 0 are  $\alpha \&\beta$ , then the value of  ${\alpha }^2+{\beta }^2$ is equal to   (where $C_r$  denotes ${ }^n{C}_r$)

${(1+x)}^n={\sum }_{r=0}^n{C}_r{x}^r$

$\Rightarrow x{(1+x)}^n={\sum }_{r=0}^n{C}_r{x}^{r+1}$

Differentiating w.r.t x we get

$xn{(1+x)}^{n-1}+{(1+x)}^n={\sum }_{r=0}^n(r+1)C_r{x}^r$

Again multiplying both sides by x

${(1+x)}^{n-1}(nx+1+x)x={\sum }_{r=0}^n(r+1)C_r{x}^{r+1}$

Again differentiating w.r.t x we get,

$\frac{d}{dx}\left({(1+x)}^{n-1}(n{x}^2+x^2+x)\right)={\sum }_{r=0}^n\left({(r+1)}^2\right)C_r{x}^r$

${\sum }_{r=0}^n\left({(r+1)}^2\right)C_r{x}^r={(1+x)}^{n-1}(2nx+2x+1)+(n{x}^2+x^2+x)(n-1){(1+x)}^{n-2}$Putting x = 1 on both sides, we get,

${\sum }_{r=0}^n\left({(r+1)}^2\right)C_r=2^{n-1}(2n+3)+(n+2).(n-1)2^{n-2}=2^{n-2}(n^2+5n+4)$Now, $f\left(x\right)=x^2+5x+4=(x+1)(x+4)\Rightarrow \alpha =-1,\beta =-4$

Hence, ${\alpha }^2+{\beta }^2=17$