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the sequence {An} is defined by A1=2 and A(n+1)=An+2n
what is the value of A100...........
9902
9900
10100
9904
Read Solution (Total 5)
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- a(n+1)-A(n)=2n
A(100)-A(99)=2*99
A(99)-A(98)=2*98
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A(2)-A(1)=2*1
finally,by addind all of these,A(100)-A(1)=2*(99+98+....+1)
=2*(99*100/2)=9900
A(100)=9900+A(1)=9902 - 12 years agoHelpfull: Yes(34) No(0)
- A1=2 ,A(n+1)=An+2n
A2=2+2*1=4(i.e: 2^2-0)
A3=4+2*2=8(i.e: 3^2-1)
A4=8+2*3=14(i.e: 4^2-2)
A5=14+2*4=22(i.e: 5^2-3)
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going so on we have A100=100^2-98=9902 - 12 years agoHelpfull: Yes(9) No(1)
- A(1)=2;
and formula is A(n+1)=A(n)+2*n;
now calculate A(2)it will be
A(2)=A(1+1)=A(1)+2*1=2+2*1=4
A(3)=A(2+1)=A(2)+2*2=4+2*2=8
A(4)=A(3+1)=A(3)+2*3=8+2*3=14
A(5)=A(4+1)=A(4)+2*4=14+2*4=22
hence the series will be
4,8,14,22
now the formula to generate this series will be..
4=(2^2)-0;
8=(3^2)-1;
14=(4^2)-2;
22=(5^2)-3;
i.e (n^2)-(n-2)
for 100
it will be
(100^100)-(100-2)
10000-98=9902... - 12 years agoHelpfull: Yes(7) No(0)
- A1=2
A2=A(1+1)=An+2n=2+2=4 or 2^2-0
A3=4+4=8 or 3^2-1
A4=8+6=14 or 4^2-2
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A100=100^2-98 = 9902 - 12 years agoHelpfull: Yes(6) No(0)
- a(n+1)-A(n)=2n
A(100)-A(99)=2*99
A(99)-A(98)=2*98
.
.
A(2)-A(1)=2*1
finally,by addind all of these,A(100)-A(1)=2*(99+98+....+1)
=2*(99*100/2)=9900
A(100)=9900+A(1)=9902 - 11 years agoHelpfull: Yes(0) No(0)
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