\(a_1=4\)
\(a_2=ra_1=4r\)
\(a_3=ra_2=r^2a_1=4r^2\)
...
\(a_9=ra_8=\cdots=r^8a_1=4r^8\)
So you have
\(1024=4r^8\implies 256=r^8\implies r=2\)
So the sum is
\(4+8+16+\cdots+512+1024\)
\(4(1+2+4+\cdots+128+256)\)
\(4(2^0+2^1+2^2+\cdots+2^7+2^8)\)
\(4\displaystyle\sum_{i=1}^92^{i-1}\)
\(4\times\dfrac{1-2^9}{1-2}\)
\(2044\)
