13,13,15,27,47,67,97,153,?
Answers
Step-by-step explanation:
from the sequence we can see that it follows the recurrence relation
Tn=2Tn−1,n>0,T0=4
now, Tn
=2Tn−1−1
=2(2Tn−2−1)−1
=22Tn−2−2−1
=22(2Tn−3−1)−2−1
=23Tn−3−(1+2+22)
.
.
.
=2n(T0)−(20+21+22+…+2n−1)
(20+21+22+…+2n−1)=2n−1,
using the formula, a(1−rn)/(1−r),
where a is the first term of the sequence(a = 1 here), r is the common ratio ( r = 2 here) , and n is the number of total terms
=2n(T0)−(2n−1)
=2n(T0–1)+1
=3∗2n+1, where T0=4 is considered
so we have derived a closed form of the given sequence, now we can determine any term in it without knowing the previous one.
you require, T7=3∗27+1=385
Answer: The answer is 385
Step-by-step explanation:
from the sequence we can see that it follows the recurrence relation
Tn=2Tn−1,n>0,T0=4
now, Tn
=2Tn−1−1
=2(2Tn−2−1)−1
=22Tn−2−2−1
=22(2Tn−3−1)−2−1
=23Tn−3−(1+2+22)
.
.
.
=2n(T0)−(20+21+22+…+2n−1)
(20+21+22+…+2n−1)=2n−1,
using the formula, a(1−rn)/(1−r),
where a is the first term of the sequence(a = 1 here), r is the common ratio ( r = 2 here) , and n is the number of total terms
=2n(T0)−(2n−1)
=2n(T0–1)+1
=3∗2n+1, where T0=4 is considered
so we have derived a closed form of the given sequence, now we can determine any term in it without knowing the previous one.
you require, T7=3∗27+1=385