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Compute the sum 1 + 4 + 7 + ... + 2005 + 2008.

 Oct 26, 2021
 #1
avatar+150 
+1

So this is a common cause of adding numbers together with large numbers (IDK what its called)
Basically, the equation is: Add the First and last term. (in your case it's 2009). Then divide by 2, and multiply by a number of numbers in that said sequence. 
So how do we how many numbers there are in the sequence? We can use the Explicit function: F(x) = 1+3 (x-1), where f(1)=1. This means that 2008-1 = 2007/3 = 669. This means there are 70 terms. So we have 2009/2) x 70. Or 2009 x 35 which is equal to 70315. 
To double-check my work, I used Wolfram Alpha! 

 Oct 26, 2021
 #3
avatar+391 
0

Would be 70280 (not 2009!) , but that's still wrong.

 

Straight

Straight  Nov 3, 2021
 #2
avatar+115420 
+2

Compute the sum 1 + 4 + 7 + ... + 2005 + 2008.

 

It is an arithmetic progression

a=1   d=3     L=2008

Tn=a+(n-1)d

2008=1+(n-1)*3

2007/3 = n-1

669=n-1

n=670

 

sn=n/2(a+L)

Sn=670/2    *(1+2008)

Sn=335  *2009

Sn=673015

 Oct 26, 2021

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