Question

What is the sum of the geometric sequence 1,-6,36... if there are 6 terms

Answer 1

$a_1=1;\ a_2=-6;\ a_3=36;\ ...\\\\r=a_2:a_1\to r=-6:1=-6\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\a_1=1;\ r=-6;\ n=6\\\\subtitute\\\\S_6=\dfrac{1(1-(-6)^6)}{1-(-6)}=\dfrac{1-46656}{1+6}=\dfrac{46655}{7}=6665$