Fill in the blanks to prove that Sn a1 1 rn1r Recall that S

Fill in the blanks to prove that S_n = a_1 (1 -r^n/1-r) Recall that S_n is the sum of the first n terms of a geometric sequence with first term a_1 and common ratio r 1. We start by multiplying both sides of S_n = a_1 (1-r^n/1-r) by (1 - r). Therefore, S_n(1 - r) = a_1(1 - r^n). By the distributive property, we know that: S_n - rS_n =. Therefore, if we can prove that S_n - rS_n =, then we will know that S_n = a_1 (1-r^n/1-r). Proof: When we write the expansions of S_n and rS_n, we see that: S_n, = a_1 + a_1r +a_1r^2 + a_1r^3...+a_1r^n-1 rS_n = From those two equations, it follows that S_n - rS_n = (a_1 + a_1r + a_1r^2 +... + a_1r^n-1)- () which can be rewritten as: S_n -rS_n = a_1 + (a_1r - a_1r) + (a_1r^2 - a_1r^2) + (a_1r^3 - a_1r^3) +... - Since every term containing r^2, where 1 lessthanorequalto k lessthanorequalto, is canceled out, we are left with: S_n -rS_n = a_1 - a_1r^2 Thus, S_n = a_1 (1 - r^n/1 - r).

Solution

By the distributive property, we know that S_n –rS_n = S_n( 1 –r) and a₁ (1 –r) = a₁ – a₁r

Therefore, if we can prove that S_n –rS_n   = a₁ – a₁r , then we will know that S_n = a₁( 1 –rⁿ/ 1- r)

Proof:

When we write the expansion of S_nand rS_n, we see that

S_n = a₁ + a₁ r + a₁ r² +a₁ r³+…+a₁ r^n-1

rS_n = a₁ r + a₁ r² + a₁ r³ + a₁ r⁴+… + a₁ rⁿ

From those two equations, it follows that S_n – rS_n = (a₁ + a₁ r + a₁ r² +a₁ r³+…+a₁ r^n-1) –

(a₁ r + a₁ r² + a₁ r³ + a₁ r⁴+… + a₁ rⁿ) which can be re-written as S_n – rS_n = a₁+( a₁r – a₁r)+(a₁r² – a₁r²) + (a₁r³ - a₁r³)+… – a₁rⁿ

Since every term containing r^k ,where 1 k n-1 is cancelled out, we are left with S_n – rS_n = a₁ – a₁rⁿ

Thus, S_n = a₁( 1 –rⁿ/ 1- r)