Question

A sequence {an} is defined recursively, with a1 = -1, and, for n > 1, an = an-1 + (-1)n. Find the first five terms of the sequence.

Answer 1

first five terms of the sequence are ; -1, 0, -1, 0, -1.

it is given that, $a_1=-1$ and n > 1 , $a_n=a_{(n-1)}+(-1)^n$

we have to find first five terms of the sequence.

putting, n = 2,

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

= $a_1+1$

= -1 + 1

= 0

putting, n = 3

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

= $a_2-1$

= 0 - 1

= -1

putting, n = 4

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

= $a_3+1$

= -1 + 1

= 0

putting, n = 5

$a_5=a_{(5-1)}+(-1)^5$

=$a_4-1$

= -1

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

A sequence is defined as below

$\displaystyle \: \sf{a_1 = - 1 } \: and \: for \: n > 1 \: , \: a_n = a_{n - 1} + {( - 1)}^{n}$

TO DETERMINE

The first five terms of the sequence.

CALCULATION

Here

$\sf{a_1 = - 1}$

$\displaystyle \: \sf{ for \: n > 1 \: , \: a_n = a_{n - 1} + {( - 1)}^{n} }$

$\therefore \: \sf{a_2 = a_1+ {( - 1)}^{2} = - 1 + 1 = 0 \: }$

$\therefore \: \sf{a_3 = a_2+ {( - 1)}^{3} = 0 - 1 = - 1\: }$

$\therefore \: \sf{a_4 = a_3+ {( - 1)}^{4} = - 1 + 1 = 0 \: }$

$\therefore \: \sf{a_5 = a_4+ {( - 1)}^{5} = 0 - 1 = - 1 \: }$

RESULT

The first five terms of the Sequence are :

-1, 0, - 1 , 0 , - 1

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