Blog: `a_n = (-1)^n(n/(n+1))` Write the first five terms of the sequence. (assume that n begins with 1.)

Sunday, May 31, 2015

$\displaystyle{a}_{{n}}={{\left(-{1}\right)}}^{{n}}{\left(\frac{{n}}{{{n}+{1}}}\right)}$ Write the first five terms of the sequence. (assume that n begins with 1.)

$\displaystyle{a}_{{n}}={{\left(-{1}\right)}}^{{n}}{\left(\frac{{n}}{{{n}+{1}}}\right)}$

$\displaystyle{a}_{{1}}={{\left(-{1}\right)}}^{{1}}{\left(\frac{{1}}{{{1}+{1}}}\right)}={\left(-{1}\right)}{\left(\frac{{1}}{{2}}\right)}=-\frac{{1}}{{2}}$

Plug in n=2, to get the 2nd term

$\displaystyle{a}_{{2}}={{\left(-{1}\right)}}^{{2}}{\left(\frac{{2}}{{{2}+{1}}}\right)}={\left(-{1}\right)}{\left(-{1}\right)}{\left(\frac{{2}}{{3}}\right)}=\frac{{2}}{{3}}$

Plug in n=3, to get the 3rd term

$\displaystyle{a}_{{3}}={{\left(-{1}\right)}}^{{3}}{\left(\frac{{3}}{{{3}+{1}}}\right)}={\left(-{1}\right)}{\left(-{1}\right)}{\left(-{1}\right)}{\left(\frac{{3}}{{4}}\right)}=-\frac{{3}}{{4}}$

Plug in n=4, to get the 4th term

$\displaystyle{a}_{{4}}={{\left(-{1}\right)}}^{{4}}{\left(\frac{{4}}{{{4}+{1}}}\right)}={\left(-{1}\right)}{\left(-{1}\right)}{\left(-{1}\right)}{\left(-{1}\right)}{\left(\frac{{4}}{{5}}\right)}=\frac{{4}}{{5}}$

Plug in n=5, to get the 5th term

$\displaystyle{a}_{{5}}={{\left(-{1}\right)}}^{{5}}{\left(\frac{{5}}{{{5}+{1}}}\right)}={\left(-{1}\right)}{\left(-{1}\right)}{\left(-{1}\right)}{\left(-{1}\right)}{\left(-{1}\right)}{\left(\frac{{5}}{{{5}+{1}}}\right)}=-\frac{{5}}{{6}}$

So, the first five terms of the sequence are $\displaystyle-\frac{{1}}{{2}},\frac{{2}}{{3}},-\frac{{3}}{{4}},\frac{{4}}{{5}},-\frac{{5}}{{6}}$

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Sunday, May 31, 2015

$\displaystyle{a}_{{n}}={{\left(-{1}\right)}}^{{n}}{\left(\frac{{n}}{{{n}+{1}}}\right)}$ Write the first five terms of the sequence. (assume that n begins with 1.)

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