If ^20C_1+(2^2)^20C_2+(3^2)^20C_3++(20^2)^20C_20=A(2^), then the ordered pair (A,) is equa…

Question

If ${}^{20}C_1+(2^2){}^{20}C_2+(3^2){}^{20}C_3+\cdots+(20^2){}^{20}C_{20}=A(2^\beta)$, then the ordered pair $(A,\beta)$ is equal to:

Accepted Answer

$$\sum_{r=1}^{20}r^2\,{}^{20}C_r = 20\sum_{r=1}^{20}r\,{}^{19}C_{r-1}=20\sum_{r=1}^{20}\left[(r-1)+1\right]{}^{19}C_{r-1}$$

$$=20\left[19\sum_{r=2}^{20}{}^{18}C_{r-2}+\sum_{r=1}^{20}{}^{19}C_{r-1}\right]=20\left[19\times2^{18}+2^{19}\right]=420\times2^{18}$$

$$(A,\beta)=(420,18)$$

Solutions · 1