Fill in the blanks to prove that Sn a11rn1r When we write t

Fill in the blanks to prove that S_n = a_1(1-r^n/1-r) 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_3r^n-1.

Solution

given Sn = a1 + a1r +a1 r^2 +a1 r^3+..............+a1 r^n-1

rSn =a1 r+ a1 r^2 +a1 r^3 +a1 r^4 +.........+a1 r^n (just multiply Sn with \'r\')

now Sn - rSn =( a1 + a1r +a1 r^2 +a1 r^3+..............+a1 r^n-1 ) - (a1 r+ a1 r^2 +a1 r^3 +a1 r^4 +.........+a1 r^n)

this can be re written as

Sn - rSn = a1 + (a1 r -a1 r ) + (a1 r^2 - a1 r^2) +.................. +(a1 r^n-1 - a1 r^n-1 ) - a1 r^n

these all terms gets canceled because r^k where 1 <= k <= n-1 (one fill in blank)

now we will be remain with

Sn - rSn = a1 - a1 r^n