Question:
If ${ }^{20} \mathrm{C}$ is the co-efficient of $\mathrm{x}^{\varepsilon}$ in the expansion of
$(1+x)^{20}$, then the value of $\sum_{r=0}^{20} r^{2}{ }^{20} C_{r}$ is equal to :
Correct Option: , 4
Solution:
$\sum_{r=0}^{20} r^{2}{ }^{20} C_{r}$
$\sum(4(\mathrm{r}-1)+\mathrm{r}){ }^{20} \mathrm{C}_{\mathrm{r}}$
$\sum r(r-1) \cdot \frac{20 \times 19}{r(r-1)} \cdot{ }^{18} C_{r}+r \cdot \frac{20}{r} \cdot \sum{ }^{19} C_{r-1}$
$\Rightarrow 20 \times 19.2^{18}+20.2^{19}$
$\Rightarrow 420 \times 2^{18}$
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