Sequences and Series · Question 48

Let an be a sequence such that a0 = 0, a1 = 12 and 2an+2 = 5an+1 - 3an, n = 0, 1, 2, 3, . Then k=1100 ak is equal to

1. 3a100 + 100
2. 3a100 - 100
3. 3a99 - 100
4. 3a99 + 100

Answer

(2) 3a100 - 100

Solution

2an+2 - 2an+1 = 3an+1 - 3an

an+2 - an+1 = 32(an+1 - an)

So the differences an+1 - an form a G.P. with first term a1 - a0 = 12 and ratio 32.

an = 12 ( 32 )n - 1 32 - 1 = ( 32 )n - 1

k=1100 ak = k=1100 ( 32 )k - 100

= 32 ( 32 )100 - 1 12 - 100

= 3 [ ( 32 )100 - 1 ] - 100

( 32 )100 - 1 = a100

k=1100 ak = 3a100 - 100

Answer: (2)