If (^30C_1)^2+2(^30C_2)^2+3(^30C_3)^2++30(^30C_30)^2=60!(30!)^2, then is equal to

Question

If $\left({}^{30}C_1\right)^2+2\left({}^{30}C_2\right)^2+3\left({}^{30}C_3\right)^2+\cdots+30\left({}^{30}C_{30}\right)^2=\dfrac{\alpha\cdot60!}{(30!)^2}$, then $\alpha$ is equal to

Accepted Answer

$$
\begin{aligned}
& \mathrm{S}=0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+1 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+2 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots \ldots+30 \cdot\left({ }^{30} \mathrm{C}_{30}\right)^2 \\
& \mathrm{S}=30 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+29 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+28 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots . \cdot+0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2 \\
& 2 \mathrm{~S}=30 \cdot\left({ }^{30} \mathrm{C}_0{ }^2+{ }^{30} \mathrm{C}_1{ }^2+\ldots \ldots . \cdot+{ }^{30} \mathrm{C}_{30}{ }^2\right) \\
& \mathrm{S}=15 \cdot{ }^{60} \mathrm{C}_{30}=15 \cdot \frac{60 !}{(30 !)^2} \\
& \frac{15 \cdot 10 !}{(30 !)^2}=\frac{\alpha \cdot 60 !}{(30 !)^2} \\
& \Rightarrow \alpha=15
\end{aligned}
$$

Other Method :

Given,

$$1\,.\,{\left( {{}^{30}{C_1}} \right)^2} + 2\,.\,{\left( {{}^{30}{C_2}} \right)^2} + 3\,.\,{\left( {{}^{30}{C_3}} \right)^2}\, + \,.....\, + \,30\,.\,{\left( {{}^{30}{C_{30}}} \right)^2}$$

$$ = \sum\limits_{r = 1}^{30} {r\,.\,{{\left( {{}^{30}{C_r}} \right)}^2}} $$

$$ = \sum\limits_{r = 1}^{30} {r\,.\,{}^{30}{C_r}\,.\,{}^{30}{C_r}} $$

$$ = \sum\limits_{r = 1}^{30} {r\,.\,{{30} \over r}\,.\,{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}} $$

$$ = 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}} $$

$$ = 30 \times $$ (Coefficient of ${x^{29}}$ in ${(1 + x)^{59}}$)

$$ = 30 \times \left( {{}^{59}{C_{29}}} \right)$$

$$ = 30 \times {{60} \over {30}} \times {}^{59}{C_{29}} \times {{30} \over {60}}$$

$$ = 30 \times {}^{60}{C_{30}} \times {{30} \over {60}}$$

$$ = 15 \times {{60!} \over {30!\,30!}}$$

$$\therefore$$ $$\alpha  = 15$$

Solutions · 1