Question

(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,.}`

Answer 1

Please see below.

Explanation:

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)`