Let $S_1=\sum _{j=1}^{10} j(j-1)\left({ }^{10}{C}_j\right)$

${\mathrm{S}}_2=\sum _{j=1}^{10} j\left({ }^{10}{C}_j\right)$ and ${\mathrm{S}}_3=\sum _{j=1}^{10} \left(j^2\right)\left({ }^{10}{C}_j\right)$

Statement-1 $S_3=55\times 2^9$
Statement-2 $S_1=90\times 2^8$ and $S_2=10\times 2^8$

Solution:

We have

$(1+x{)}^{10}=\sum _{j=0}^{10} \left({ }^{10}{C}_j\right)x^j$

Differentiating both the sides with respect to x, we get

$10(1+x{)}^9=\sum _{j=1}^{10} j\left({ }^{10}{C}_j\right)x^{j-1}$                                   (1)

Again differentiating both the sides with respect to x, we get

$\left(10\right)\left(9\right)(1+x{)}^8=\sum _{j=1}^{10} j(j-1)\left({ }^{10}{C}_j\right)x^{j-2}$                   (2)

Putting $x=1$ in (1) and (2), we get

$S_2=\sum _{j=1}^{10} j\left({ }^{10}{C}_j\right)=10\left(2^9\right)$

and $S_1=\sum _{j=1}^{10} j(j-1)\left({ }^{10}{C}_j\right)=\left(10\right)\left(9\right)\left(2^8\right)=\left(90\right)\left(2^8\right)$

Adding the above two equation, we get

$\begin{array}{l}{S}_3=\sum _{j=1}^{10} [j+j(j-1\left)\right]\left({ }^{10}{C}_j\right)\\ =\left(10\right)\left(2^8\right)(2+9)\\ \Rightarrow S_3=\left(55\right)\left(2^9\right)\end{array}$

Thus, Statement-1 is true but Statement-2 is false.