Direction: Write the 1^("th ")4 terns of the sequence whose nth term is given by the formula a_(n)=n+1 a_(n)=2-2n (2) 345 x=2-2(1)=0 a_(2)=2-2(2)=-2 0-2-4-6dots a_(3)=2-2(3)=-4 a_(n)=2*2(4)=a_(n)=2^(n) a_(n)=n-1 a_(n)=2^(n) a_(n)=1-1=0 a_(1)= a_(2)=2+1=1quada_(1)∣,2,3//dots a_(2)=0,2= 2,48,16,dots a_(4)=4-1=3 a_(2)=2^(3)=8 a_(n)=2n+1 a_(n)=n^(2)+1 a_(1) 2(n)-1=3 a_(2)2(2)+1=5,3,5,7,9dots a_(3):2(3)+1=7 a_(4)=2(m)+1=9 a_(1)=1^(2)+1=2 a_(2)=2^(2)+1=5quad2","5","10","17","dots a_(3)=3^(2)+1=10 a_(4)=4^(8)+1=17 nn a_(4)=4^(2)+1=17 n 7. a_(n)=3n-1 8. a_(n)= bar(n+1) a_(1)=-3(1)=1=2 a_(3)=3(3)-1=8 2,5,8,11,km a_(1)=(1)/(1+1)=(1)/(2) 94=(4)/(4+1)=(9)/(5) a_(4)=3(4)-1=11 a_(2)=(2)/(2+1)=(2)/(3) (1)/(2),(2)/(3),(3)/(4),(4)/(5),dots 9. a_(n)=1-2n 10. a_(n)=n-(1)/(n)

In this question, it’s asked that the first four terms of the sequence for each of the formulae mentioned.1. In the first part, the sequence is derived from formula . Thus for n values of 0 through 3, the sequence turns out to be 1, 2, 3 and 4. 2. In the second scenario, the formula is . When n takes values 1 through 4, the sequence is found to be 0, -2, -4 and -6. 3. For the third case, the nth term is governed by . Thus, we acquire the sequence 0, 1, 2 and 3 for initial four terms. 4. Here, leads to the sequence 2, 4, 8 and 16 for n=1 to n=4.5. When governed by the principle , the identified sequence turned out to be 2, 5, 10, and 17 after substituting values of n 1 through 4.6. In case of \(a_{n}=n/(n+1)\), the sequence identified from first to the fourth term is 1/2, 2/3, 3/4, and 4/5.7. Finally, when , the initial four terms comprise the sequence 0, -1/2, -2/3, -3/4.