Question

What is the sum of the n terms of the series 6 + 24 + 96 + 384 + . . . ? Hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r ≠ 1, where a1 is the first term and r is the common ratio. Sn = −6(4n − 1) Sn = 6(1 − 4n) Sn = 2(4n − 1) cap s sub n equals start fraction six left parenthesis four to the power of n end power minus one right parenthesis over n minus one end fraction

Answer 1

Note that each term is 6 times a power of 4, starting from 0. So you can write

$S_n=6(1+4+4^2+4^3+\cdots+4^{n-1})$

Multiplying both sides by 4 gives

$4S_n=6(4+4^2+4^3+4^4+\cdots+4^n)$

Subtracting this from the first sum reduces to

$S_n-4S_n=6(1-4^n)$
$-3S_n=6(1-4^n)$
$S_n=2(4^n-1)$