2016 adv paper q15 (1 Viewer)

[waltssillyhat]

Can someone please explain how the worked solution got (7/8)^n-1 for part (ii)? If we are finding the geometric sum, wouldn't r = 7/8?

 

[yourlocaliga]

im pre sure they do use r=7/8, with n-1 terms, as you have all the the sum of 1/8*(7/8) and whatnot n-2 times, so thats n-2 terms plus the initial 1/8 which makes n-1 terms

they just use the other variation of the geometric sum on the formula sheet where everythings the other way around

 

[C2H6O]

Can someone please explain how the worked solution got (7/8)^n-1 for part (ii)? If we are finding the geometric sum, wouldn't r = 7/8?
View attachment 48294View attachment 48295

yea they did put r=7/8 right? this is then raised to the number of terms which is if you look at it as starting from (7/8)^0 all the way to (7/8)^(n-2) inclusive, thats n-1 terms. then theyve multiplied the 1/8 into the fraction which makes the denominator -1, which is distributed into the brackets in the numerator