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

## 1(a) $f \left(n\right) = {\left(- 1\right)}^{n} \frac{2 n + 1}{n} ^ 2$ 1(b) $f \left(n\right) = \sin \frac{\frac{3 \pi}{n}}{n}$ 2(a) $\left\{1 , \frac{1}{2} , \frac{1}{6} , \frac{1}{24} , \frac{1}{120} , .\right\}$ 2(b) $\left\{3 , \frac{3}{2} , 1 , \frac{3}{4} , \frac{3}{5} , .\right\}$ 2(c) $\left\{1 , - \sqrt{7} , 7 , - 7 \sqrt{7} , .\right\}$

Mar 7, 2017

#### Explanation:

1(a) As $f \left(n\right) = {\left(- 1\right)}^{n} \frac{2 n + 1}{n} ^ 2$, we can get firsr five terms by putting $n = 1 , 2 , 3 , 4$ and $5$ and we get

$\left\{\frac{\left(- 1\right) \times 3}{1} ^ 2 , \frac{{\left(- 1\right)}^{2} \times 5}{2} ^ 2 , \frac{{\left(- 1\right)}^{3} \times 7}{3} ^ 2 , \frac{{\left(- 1\right)}^{4} \times 9}{4} ^ 2 , \frac{{\left(- 1\right)}^{5} \times 11}{5} ^ 2\right\}$

or $\left\{- 3 , \frac{5}{4} , - \frac{7}{9} , \frac{9}{16} , - \frac{11}{25}\right\}$

1(b) As $f \left(n\right) = \sin \frac{\frac{3 \pi}{n}}{n}$, we can get firsr five terms by putting $n = 1 , 2 , 3 , 4$ and $5$ and as $\pi$ radian is ${180}^{\circ}$, we get

$\left\{\sin {540}^{\circ} / 1 , \sin {270}^{\circ} / 2 , \sin {180}^{\circ} / 3 , \sin {135}^{\circ} / 4 , \sin {108}^{\circ} / 5\right\}$

or $\left\{0 , - \frac{1}{2} , 0 , \frac{1}{4 \sqrt{2}} , \frac{\sqrt{10 + 2 \sqrt{5}}}{20}\right\}$

2(a) We have first five terms as $\left\{1 , \frac{1}{2} , \frac{1}{6} , \frac{1}{24} , \frac{1}{120} , .\right\}$, 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 $\left\{3 , \frac{3}{2} , 1 , \frac{3}{4} , \frac{3}{5} , .\right\}$, which can be written as $\left\{\frac{3}{1} , \frac{3}{2} , \frac{3}{3} , \frac{3}{4} , \frac{3}{5} , . .\right\}$.

Hence formula is $f \left(n\right) = \frac{3}{n}$

2(c) We have first five terms as $\left\{1 , - \sqrt{7} , 7 , - 7 \sqrt{7} , .\right\}$, where ratio between a term and its immediately preceding term is constant as

$\frac{- \sqrt{7}}{1} = \frac{7}{- \sqrt{7}} = \frac{- 7 \sqrt{7}}{7} = - \sqrt{7}$.

Therefore, it is a geometric sequence with first term as $1$ and common ratio as $- \sqrt{7}$.

Hence formula is $f \left(n\right) = 1 \times {\left(- \sqrt{7}\right)}^{n - 1} = {\left(- \sqrt{7}\right)}^{n - 1}$