Question 1167598
(a) {{{b[n] = c*5^n + d*(-4)^n}}} for {{{n>=0}}} 
means that {{{b[0] = c*5^0 + d*(-4)^0=c+d=0}}}, by hypothesis, after direct substitution.
Also, inserting {{{n=1}}} into the formula gives {{{b[1] = c*5^1 + d*(-4)^1=-9}}}.
Hence, {{{5c-4d=-9}}}.

From c + d = 0, we get d = -c. 
Substituting this into the equation 5c - 4d = -9 gives 5c - 4(-c) = 5c+4c = -9,
or 9c = -9, or c = -1. 
Therefore, c = -1, and d = -(-1) = 1.

(b) Similarly, {{{b[n] = c*5^n + d*(-4)^n}}} gives {{{b[0] = c*5^0 + d*(-4)^0=c+d=4}}}, or c + d = 4, or d = 4 - c. Also, {{{b[1] = c*5^1 + d*(-4)^1=11}}}, or 5c - 4d = 11.

Substituting, 5c - 4(4 - c) = 11, or 5c - 16 + 4c = 11.
This gives 9c = 27, or c = 3, which consequently gives d = 4 - 3 = 1.

(c) From part (b), we get  {{{b[n] = 3*5^n + (-4)^n}}}, so that {{{b[2] = 3*5^2 + (-4)^2=3*25+16 = 75+16=91}}}