Find the value of tg of a sequence, given
that tn = tn-1 + 2" and t1 = 1.
Answers
Answer:
Correct option is
B
(2017)×(2015)×(2013)×....×(1)
(2018)×(2016)×(2014)×....×(2)
T
n+2
=T
n
+
T
n+1
1
⇒T
n+2
−T
n
=
T
n+1
1
⇒T
n+2
⋅T
n+1
−T
n+1
⋅T
n
=1
Now, let a
n
=T
n+1
⋅T
n
⇒a
n+1
−a
n
=1
So, a
1
,a
2
,a
3
,...,a
n+1
form an A.P. with common difference, d=1; and first term a
1
=T
2
⋅T
1
=(1)(1)=1
Now, n
th
term of this AP is a
n
=a
1
+(n−1)d
a
n
=1+(n−1)1
a
n
=n
So, T
n+1
⋅T
n
=n
So, (T
2019
)(T
2018
)=2018 and (T
2018
)(T
2017
)=2017
⇒T
2019
=
T
2018
2018
=
(
T
2017
2017
)
2018
−(
2017
2018
)T
2017
Thus, T
2019
=(
2017
2018
)T
2017
Similarly, T
2017
=(
2015
2016
)T
2015
T
2015
=(
2013
2014
)T
2013
T
3
=(
1
2
)T
2
=2
Thus, T
2019
=(
2017
2018
)(
2015
2016
)(
2013
2014
)(
1
2
)
So T
2019
=
(2017)(2015)(2013).(3)(1)
(2018)(2016)(2014)(2)
So option B.
Step-by-step explanation: