The recursive sequence is defined by the formula #t_n=2t_(n-1)+3#; and #t_1=-2#, how do you find #t_6#?

2 Answers
May 26, 2016

Answer:

Find #t_n = 2^(n-1)-3# and hence #t_6 = 2^5-3 = 29#

Explanation:

Note that as #n# and #t_n# get larger, the sequence will approximate a geometric sequence with common ratio #2#.

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

#t_n = 2^na + b#

for some constants #a# and #b#.

Then we find:

#2^na + b = t_n = 2t_(n-1)+3 = 2(2^(n-1)a + b) = 2^na+2b+3#

Hence we find #b = -3#

Then:

#-2 = t_1 = 2^1 a - 3 =2a-3#

Hence we find #a = 1/2#

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

#t_n = 2^(n-1)-3#

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_n#

In particular:

#t_6 = 2^5-3 = 32-3 = 29#

May 26, 2016

Answer:

#t_6=2^5-3#

Explanation:

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

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

#t_n = c_0 2^n-3# and also
#t_1=c_0 2^1-3->c_0=1/2#. Putting all together
#t_n=2^{n-1}-3->t_6=2^5-3#