Suppose that Sn = 3 + 1/3 + 1/9 + ... + 1/3(n-2)

[math_celebrity]

Suppose that Sn = 3 + 1/3 + 1/9 + ... + 1/3(n-2)

a) Find S10 and S∞
b) If the common difference in an arithmetic sequence is twice the first term, show that Sn/Sm = n^2/m^2

a) Sum of the geometric sequence is
a = 3 and r = 1/3

(a(1 - r)^n)/(1 - r)

(3(1 - 1/3)^9)/(1 - 1/3)

S10 = 4.499771376

For infinity, as n goes to infinity, the numerator goes to 1
so we have S∞ = 3(1)/2/3 = 4.5

b) Sum of an arithmetic sequence formula is below:

n(a1 + an)/2

an = a1 + (n - 1)2a1 since d = 2a1

n(a1 + a1 + (n - 1)2a1)/2

(2a1n + n^2 - 2a1n)/2

n^2/2

For Sm
m(a1 + am)/2

am = a1 + (m - 1)2a1 since d = 2a1

m(a1 + 1 + (m - 1)2a1)/2

(2a1m + m^2 - 2a1m)/2

m^2/2

Sn/Sm = n^2/m^2 (cancel the 2's)

S10/S1 = 10^2/1^2

We know S₁ = 3

So we have 100(3)/1

S10 = 300