How do you find $a_{16}$ $a_16$ given $a_{n} = {(- 1)}^{n - 1} [n (n - 1)]$ $a_n=(-1)^(n-1)[n(n-1)]$ ?

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How do you find $a_{16}$ $a_16$ given $a_{n} = {(- 1)}^{n - 1} [n (n - 1)]$ $a_n=(-1)^(n-1)[n(n-1)]$ ?

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Answer 1 · 2017-12-29T19:00:46

$a_{16} = - 240$ $a_(16)=-240$

Explanation:

$substitute n =16 into the nth term formula$ $"substitute n =16 into the nth term formula"$

$\Rightarrow a_{16} = {(- 1)}^{15} [16 \times 15]$ $rArra_(16)=(-1)^(15)[16xx15]$

$\times \times x = - 1 \times 240 = - 240$ $color(white)(xxxxx)=-1xx240=-240$

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How do you find $a_{16}$ $a_16$ given $a_{n} = {(- 1)}^{n - 1} [n (n - 1)]$ $a_n=(-1)^(n-1)[n(n-1)]$ ?

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How do you find a16a_16 given an=(−1)n−1[n(n−1)]a_n=(-1)^(n-1)[n(n-1)]?

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How do you find $a_{16}$ $a_16$ given $a_{n} = {(- 1)}^{n - 1} [n (n - 1)]$ $a_n=(-1)^(n-1)[n(n-1)]$ ?