If a(1)=1 and a(n+1)-3a(n)+2=4n for every positive integer n,then a(100) equals ?

Question

If a(1)=1 and a(n+1)-3a(n)+2=4n for every positive integer n,then a(100) equals ?
(a) (3 raised to the power 99)-200
(b) (3 raised to the power 99)+200
(c) (3 raised to the power 100)-200
(d) (3 raised to the power 100)+200

Answer 1

It's c: a(100) = 3^(100) - 2(100)

Given a(n + 1) - 3a(n) + 2 = 4n, then
a(n + 1) = 4n + 3a(n) - 2

If a(1) = 1, then
for n = 1, a(2) = 4(1) + 3(1) - 2 = 4 + 3 - 2 = 5;
for n = 2, a(3) = 4(2) + 3(5) - 2 = 8 + 15 - 2 = 21;
for n = 3, a(4) = 4(3) + 3(21) - 2 = 12 + 63 - 2 = 73;
for n = 4, a(5) = 4(4) + 3(73) - 2 = 16 + 219 - 2 = 233.

a(2) = 3^2 - 2(2) = 9 - 5 = 4.
a(3) = 3^3 - 2(3) = 27 - 6 = 21.
a(4) = 3^4 - 2(4) = 81 - 8 = 73.
a(5) = 3^5 - 2(5) = 243 - 10 = 233.

Generalizing, is a(n) = 3^n - 2n ?

Is a(1) true?
a(1) = 3^(1) - 2(1) = 3 - 2 = 1.
a(1) is true.

Assuming a(k) = 3^k - 2k is true, is a(k + 1) true?
a(k + 1) = 4k + 3a(k) - 2 [from given statement]
a(k + 1) = 4k + 3(3^k - 2k) - 2
a(k + 1) = 3k + k + 3^(k + 1) - 6k - 2
a(k + 1) = 3^(k + 1) - 2k - 2
a(k + 1) = 3^(k + 1) - 2(k + 1)
a(k + 1) is true.

Therefore, a(n) = 3^n - 2n,
and a(100) = 3^(100) - 2(100).

Answer 2

b)