The given formula means that, after the first two terms of the sequence, every term a_(n+2) is the difference between two previous terms a_n and a_(n+1), both of which are multiplied by 4 and 2 respectively.
a_1&=4
a_2&=5
a_(n+2)&=4a_n-2a_(n+1)
To do so, we will use a table. To find a_3, we will substitute 1 for n in the above formula. To find a_4, we will substitute 2 for n, and so on.
n
a_(n+2)=4a_n-2a_(n+1)
4a_n-2a_(n+1)
a_(n+2)
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a_1= 4
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a_2= 5
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1
a_(1+2)=4a_1-2a_(1+1) ⇕ a_3=4a_1-2a_2
a_3=4 a_1-2 a_2 ⇓ a_3=4( 4)-2( 5)
a_3= 6
2
a_(2+2)=4a_2-2a_(2+1) ⇕ a_4=4a_2-2a_3
a_4=4 a_2-2 a_3 ⇓ a_4=4( 5)-2( 6)
a_4= 8
3
a_(3+2)=4a_3-2a_(3+1) ⇕ a_5=4a_3-2a_4
a_5=4 a_3-2 a_4 ⇓ a_5=4( 6)-2( 8)
a_5= 8
The first five terms of the sequence are 4, 5, 6, 8, and 8.
