How do you find the first five terms given $a_{1} = 2$ $a_1=2$ and $a_{n + 1} = a_{n} - n$ $a_(n+1)=a_n-n$ ?

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Answer 1 · 2016-11-03T05:03:39

First five terms are ${2, 1, - 1, - 4, - 8}$ ${2,1,-1,-4,-8}$

$a_{1} = 2$ $a_1=2$

Now as $a_{n + 1} = a_{n} - n$ $a_(n+1)=a_n-n$

$a_{2} = a_{1 + 1} = a_{1} - 1 = 2 - 1 = 1$ $a_2=a_(1+1)=a_1-1=2-1=1$

$a_{3} = a_{2 + 1} = a_{2} - 2 = 1 - 2 = - 1$ $a_3=a_(2+1)=a_2-2=1-2=-1$

$a_{4} = a_{3 + 1} = a_{3} - 3 = - 1 - 3 = - 4$ $a_4=a_(3+1)=a_3-3=-1-3=-4$

$a_{5} = a_{4 + 1} = a_{4} - 4 = - 4 - 4 = - 8$ $a_5=a_(4+1)=a_4-4=-4-4=-8$

Hence first five terms are ${2, 1, - 1, - 4, - 8}$ ${2,1,-1,-4,-8}$

How do you find the first five terms given a1=2a_1=2 and an+1=an−na_(n+1)=a_n-n?