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 $\alpha$  and $\beta ,$  then the value of ${\alpha }^2+{\beta }^2$  is equal to (where
Cr denotes ${ }^n{C}_r$ )

 ${(1+x)}^n={\sum }_{r=0}^n{C}_r{x}^r$
 $=x{(1+x)}^n={\sum }_{r=0}^n{x}^{r+1}$
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,
 $\frac{d}{x}{\left((1+x)\right)}^{n-1}(n{x}^2+x^2+x)x={\sum }_{r=0}^n\left({(r+1)}^2\right)C_r{x}^r$
 ${\sum }_{r=0}^n\left((r+1^2)\right)C_r{x}^r={(1+x)}^{n-1}(2nx+2x+1)+(n{x}^2+x^2+x)(n-1){(1+x)}^{n-2}$
Putting $x=1$ 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,  ${\alpha }^2+{\beta }^2=17$