Question
Mathematics
Write the first four terms of the sequence with the given rule.
1. ${a}_{n}=2n+3$a_(n)=2n+3
2. ${a}_{n}=3n-7$a_(n)=3n-7
3. ${a}_{n}=5n+6$a_(n)=5n+6
4. ${a}_{n}=4n+5$a_(n)=4n+5
5. ${a}_{n}=2\left(n+3\right)$a_(n)=2(n+3)
6. ${a}_{n}=\sqrt{{n}^{2}}$a_(n)=sqrt(n^(2))
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1. ${a}_{1}=2\left(1\right)+3=5,{a}_{2}=2\left(2\right)+3=7,{a}_{3}=2\left(3\right)+3=9,{a}_{4}=2\left(4\right)+3=11$a_(1)=2(1)+3=5,a_(2)=2(2)+3=7,a_(3)=2(3)+3=9,a_(4)=2(4)+3=11
The first four terms are $5,7,9,11$5,7,9,11.
2. ${a}_{1}=3\left(1\right)-7=-4,{a}_{2}=3\left(2\right)-7=-1,{a}_{3}=3\left(3\right)-7=2,{a}_{4}=3\left(4\right)-7=5$a_(1)=3(1)-7=-4,a_(2)=3(2)-7=-1,a_(3)=3(3)-7=2,a_(4)=3(4)-7=5
The first four terms are $-4,-1,2,5$-4,-1,2,5.
3. ${a}_{1}=5\left(1\right)+6=11,{a}_{2}=5\left(2\right)+6=16,{a}_{3}=5\left(3\right)+6=21,{a}_{4}=5\left(4\right)+6=26$a_(1)=5(1)+6=11,a_(2)=5(2)+6=16,a_(3)=5(3)+6=21,a_(4)=5(4)+6=26
The first four terms are $11,16,21,26$11,16,21,26.
4. ${a}_{1}=4\left(1\right)+5=9,{a}_{2}=4\left(2\right)+5=13,{a}_{3}=4\left(3\right)+5=17,{a}_{4}=4\left(4\right)+5=21$a_(1)=4(1)+5=9,a_(2)=4(2)+5=13,a_(3)=4(3)+5=17,a_(4)=4(4)+5=21
The first four terms are $9,13,17,21$9,13,17,21.
5. ${a}_{1}=2\left(1+3\right)=8,{a}_{2}=2\left(2+3\right)=10,{a}_{3}=2\left(3+3\right)=12,{a}_{4}=2\left(4+3\right)=14$a_(1)=2(1+3)=8,a_(2)=2(2+3)=10,a_(3)=2(3+3)=12,a_(4)=2(4+3)=14
The first four terms are $8,10,12,14$8,10,12,14.
6. ${a}_{1}=\sqrt{{1}^{2}}=1,{a}_{2}=\sqrt{{2}^{2}}=2,{a}_{3}=\sqrt{{3}^{2}}=3,{a}_{4}=\sqrt{{4}^{2}}=4$a_(1)=sqrt(1^(2))=1,a_(2)=sqrt(2^(2))=2,a_(3)=sqrt(3^(2))=3,a_(4)=sqrt(4^(2))=4
The first four terms are $1,2,3,4$1,2,3,4.
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