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The sum of the first n terms of a sequence is n^2 + 5n.  What is the 1000th term of the sequence?

 May 7, 2020
 #1
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The list of numbers is just even numbers starting at 6, so the 1000th term would be 2006.

 May 7, 2020
 #2
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The sequence is:
3, 6, 9, 12, 15, 18, 21 =n^2 + 5*n=7^2 + 35=84
Therefore the 1000th term =First term + 3*(1000 - 1)
                                                  =3 + (3*999)
                                                   =3 + 2,997
                                                   =3,000 - which is 1000th term.

 May 7, 2020
edited by Guest  May 7, 2020
 #3
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The sum of the first term is just the first term. So if we plug in 1 to the equation, we'll find the first term which is (1^2) + 5 * 1 = 6. From this I can already see the difference in our solutions.

HELPMEEEEEEEEEEEEE  May 7, 2020
 #4
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But, doesn't he/she say "The sum of first n terms" =n^2 + 5n, which implies "several terms" summing up to n^2 + 5*n ?? Otherwise, he/she could have said " The sum of the FIRST term" =n^2 + 5n.

Guest May 7, 2020
 #5
avatar+26398 
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The sum of the first n terms of a sequence is n2+5n.  
What is the 1000th term of the sequence?

 

Formula:  sn=(a1+an)2n

 

a1=s1|sn=n2+5n=12+51a1=6

 

sn=(a1+an)2n|a1=6, sn=n2+5nn2+5n=(6+an)2n2n2+10n=(6+an)n|:n2n+10=6+an2n+4=anan=2n+4

 

an=2n+4|n=1000a1000=21000+4a1000=2004

 

laugh

 May 8, 2020

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