ind the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, .

Required sum $=2 \times 128+4 \times 32+8 \times 8+16 \times 2+32 \times \frac{1}{2}$

$=64\left[4+2+1+\frac{1}{2}+\frac{1}{2^{2}}\right]$

Here, $4,2,1, \frac{1}{2}, \frac{1}{2^{2}}$ is a G.P.

First term, $a=4$

Common ratio, $r=\frac{1}{2}$

It is known that, $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$

$\therefore S_{5}=\frac{4\left[1-\left(\frac{1}{2}\right)^{5}\right]}{1-\frac{1}{2}}=\frac{4\left[1-\frac{1}{32}\right]}{\frac{1}{2}}=8\left(\frac{32-1}{32}\right)=\frac{31}{4}$

$\therefore$ Required sum $=64\left(\frac{31}{4}\right)=(16)(31)=496$