Thread: geometric sequences

show why
(r^(n+1)-1)/(r-1) = the sum (between k=0 and n) of r^k = Sn (The n in Sn is lower than the S)

he hints to multiply by r-1 or use long division

then I'm supposed to use part to show why
the sum K=0 to infinity = (1)/(1-r)

i can graph the sequences and I know its true but I don't understand what to do tomake it work.

thankyou,
Cailtyn

Originally Posted by horsejumper

show why
(r^(n+1)-1)/(r-1) = the sum (between k=0 and n) of r^k = Sn (The n in Sn is lower than the S)

he hints to multiply by r-1 or use long division

(sum_{k=0 to n} r^k)(r-1) = sum_{k=0 to n} [r^{k+1} - r^k]

.....................................= (r-1) + (r^2-r) + ... + (r^n-r^{n-1)) + (r^{n+1} - r^n)

Now all but the -1 in the first term and the r^{n+1} in the last term cancel
so:

(sum_{k=0 to n} r^k)(r-1) = [r^{n+1}-1]

or:

sum_{k=0 to n} r^k = [r^{n+1}-1]/(r-1).

then I'm supposed to use part to show why
the sum K=0 to infinity = (1)/(1-r)

From the first part we have:

(sum_{k=0 to n} r^k) = [r^{n+1}-1]/(r-1)

Now if |r|<1

lim_{n->infty} (sum_{k=0 to n} r^k) = (sum_{k=0 to infty} r^k) = -1/(r-1) = 1/(1-r)

as lim_{n->infty} r^{n+1} = 0.

RonL

Thankyou soooo much!!!!!!!!!!!!!!!!!!!!!!!!!