The recursive sequence is defined by the formula t_n=2t_(n-1)+3tn=2tn1+3; and t_1=-2t1=2, how do you find t_6t6?

2 Answers
May 26, 2016

Find t_n = 2^(n-1)-3tn=2n13 and hence t_6 = 2^5-3 = 29t6=253=29

Explanation:

Note that as nn and t_ntn get larger, the sequence will approximate a geometric sequence with common ratio 22.

So we can look for a general formula for terms of the form:

t_n = 2^na + btn=2na+b

for some constants aa and bb.

Then we find:

2^na + b = t_n = 2t_(n-1)+3 = 2(2^(n-1)a + b) = 2^na+2b+32na+b=tn=2tn1+3=2(2n1a+b)=2na+2b+3

Hence we find b = -3b=3

Then:

-2 = t_1 = 2^1 a - 3 =2a-32=t1=21a3=2a3

Hence we find a = 1/2a=12

So the general formula of a term of our sequence is:

t_n = 2^(n-1)-3tn=2n13

If you like, we can double check this formula:

2t_(n-1)+3 = 2(2^(n-2)-3)+3 = 2^(n-1)-6+3 = 2^(n-1)-3 = t_n2tn1+3=2(2n23)+3=2n16+3=2n13=tn

In particular:

t_6 = 2^5-3 = 32-3 = 29t6=253=323=29

May 26, 2016

t_6=2^5-3t6=253

Explanation:

Proposing t_n = c_0 a^n+btn=c0an+b and substituting we get
c_0a^n+b = 2(c_0a^{n-1}+b)+3c0an+b=2(c0an1+b)+3 or
c_0a^n-2c_0a^{n-1}=2b-b+3c0an2c0an1=2bb+3.

Making a = 2a=2 and b = -3b=3 the equality is observed. So we get

t_n = c_0 2^n-3tn=c02n3 and also
t_1=c_0 2^1-3->c_0=1/2t1=c0213c0=12. Putting all together
t_n=2^{n-1}-3->t_6=2^5-3tn=2n13t6=253