Please help me with my math problem?

Question

Consider the sequences,An =(-2/5)^Bn,Cn =(2/5)^Bn, where the sequence, Bn = 3n-1, n=0,1,2....
(a) Find the quantity, Q = sum( Bn,n=0,30 ).
(b) Find the quantities, S = sum( (An/Cn),n=0,50 ) and T = sum( An,n=0,25 ).
(c) Let Wn = sum( (Ai*Ci), i=0, n )Find the long term behavior of the sequence,Wn

Answer 1

Q = sum(Bn) = 3*sum(n) - sum(1)
The first term is an arithmetic series with first term 0 and last term 30 (31 terms in all). The second term is just adding 1 31 times.
Q = 3 * 31/2 *(0 + 30) - 31 = 1364

S = An/Cn = sum[ (-1)^(3n-1) ]
Since 3n-1 is odd for n = even
and even for n = odd
S = -1+1-1+1 + ... -1
= -1
The last term is -1 and all other terms sum to zero

T = sum(An) = (-2/5)^-1 + (-2/5)^2 + (-2/5)^5 + ...
This is a geometric series of 26 terms with
a = -5/2, r = -8/125
T = -2.3496

Wn = sum(-4/25)^(3i-1)
AS i goes to infinity, the terms in the series go to zero, so Wn which is the sum of all term up to n, goes to a constant as n goes to infinity.

Answer 2

the answer is sum=

Answer 3

(a) Q=13,764