## College Algebra 7th Edition

$P=10^{19}$
We are asked to find the product: $P=10^{1/10}\cdot 10^{2/10}\cdot 10^{3/10}\cdot 10^{4/10}... 10^{19/10}$ We simplify the powers: $10^{(1+2+3+4+...+19)/10}$ We see that we have an arithmetic sequence with $a=1$ and $a_{19}=19$. We find the partial sum: $S_{19}=\frac{19}{2}(1+19)=19*20/2=19*10=190$ Thus the product is: $P=10^{190/10}=10^{19}$