(1) Find first five terms f the following sequences? (2) Given first few terms find the explicit formula for #n^(th)# term?

Question

1(a) #f(n)=(-1)^n(2n+1)/n^2#
1(b) #f(n)=sin((3pi)/n)/n#
2(a) #{1,1/2,1/6,1/24,1/120,.}#
2(b) #{3,3/2,1,3/4,3/5,.}#
2(c) #{1,-sqrt7,7,-7sqrt7,.}#

Julia Adams · Accepted Answer

undefined (1) The first five terms of a sequence are the first five numbers in the sequence.
(2) To find the explicit formula for the nth term of a sequence, you can use the method of finite differences or observe the pattern in the given terms to derive a formula that relates the term number (n) to the value of the term.

Hazel Anglin · Answer

Please see below.

1(a) As #f(n)=(-1)^n(2n+1)/n^2#, we can get firsr five terms by putting #n=1,2,3,4# and #5# and we get #{((-1)xx3)/1^2,((-1)^2xx5)/2^2,((-1)^3xx7)/3^2,((-1)^4xx9)/4^2,((-1)^5xx11)/5^2}# or #{-3,5/4,-7/9,9/16,-11/25}# 1(b) As #f(n)=sin((3pi)/n)/n#, we can get firsr five terms by putting #n=1,2,3,4# and #5# and as #pi# radian is #180^@#, we get #{sin540^@/1,sin270^@/2,sin180^@/3,sin135^@/4,sin108^@/5}# or #{0,-1/2,0,1/(4sqrt2),sqrt(10+2sqrt5)/20}# 2(a) We have first five terms as #{1,1/2,1/6,1/24,1/120,.}#, which can be written as #{1/(1!),1/(2!),1/(3!),1/(4!),1/(5!),..}#. Hence formula is #f(n)=1/(n!)# 2(b) We have first five terms as #{3,3/2,1,3/4,3/5,.}#, which can be written as #{3/1,3/2,3/3,3/4,3/5,..}#. Hence formula is #f(n)=3/n# 2(c) We have first five terms as #{1,-sqrt7,7,-7sqrt7,.}#, where ratio between a term and its immediately preceding term is constant as #(-sqrt7)/1=7/(-sqrt7)=(-7sqrt7)/7=-sqrt7#. Therefore, it is a geometric sequence with first term as #1# and common ratio as #-sqrt7#. Hence formula is #f(n)=1xx(-sqrt7)^(n-1)=(-sqrt7)^(n-1)#

(1) Find first five terms f the following sequences? (2) Given first few terms find the explicit formula for #n^(th)# term?

1(a) #f(n)=(-1)^n(2n+1)/n^2# 1(b) #f(n)=sin((3pi)/n)/n# 2(a) #{1,1/2,1/6,1/24,1/120,.}# 2(b) #{3,3/2,1,3/4,3/5,.}# 2(c) #{1,-sqrt7,7,-7sqrt7,.}#

1(a) #f(n)=(-1)^n(2n+1)/n^2#
1(b) #f(n)=sin((3pi)/n)/n#
2(a) #{1,1/2,1/6,1/24,1/120,.}#
2(b) #{3,3/2,1,3/4,3/5,.}#
2(c) #{1,-sqrt7,7,-7sqrt7,.}#