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Question

26. fill in the banks to prove that Sn =a1(1-rn/1-r )
PLAN: Recall that Sn is the sum of the first terms of a geometric sequence with first term a1 and common ratio r not = to 1. We start by multiplying both sides of
Sn =a1(1-rn/1-r ) by (1-r). Therefore, Sn(1-r) = a1(1-rn). By the distributive property, we know that: Sn-rSn=_____. Therefore, if we can prove that Sn-rSn=_____, then we will know that Sn =a1(1-rn/1-r ).
PROOF: When we write the expansions Sn and rSn, we see that: Sn =a1+a1r+a1r2+a1r3+... a1rn-1
rSn=_______.
From those two equations, it follows that Sn-rSn= (a1+a1r+a1r2+a1r3+... a1rn-1) -(_______) which can be written as: Sn-rSn=a1+(a1r-ar^2)+(a1r^3-a1r^3)+...-_____
Since every term containing r^k, where 1 <than or equal to k < than or equal to____, Is canceled out, we are left with: Sn-rSn=a1-a1r^n
Thus, Sn =a1(1-rn/1-r ).

Answer 1

http://www.mathalino.com/reviewer/derivation-of-formulas/sum-of-finite-and-infinite-geometric-progression

 

That is a link to the exact same proof.

 

 They multiply both sides of the expansion by (1-r)

 

Upon distribution, the cross terms cancel out.

 

You have to read the proof thoroughly to understand it.