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Gauss's Problem and Arithmetic Series

Let's look at an easier example to help us figure this out:

the summation of ( 3 + 2k ) as k goes from 5 to 7 = 13 + 15 + 17 ... 13 is when k = 5 , 15 is when k = 6 , 17 is when k = 7

There are THREE terms.

terms: a1 , a2 , a3 , a4 , a5 , a6 , a7 ... the first 4 terms get lopped off!

             So, it's 7 - 4:

the summation as k goes from 5 to 7 ... the number of terms is 7 ( the guy on top of sigma ) minus one less than 4 ( the guy k equals underneath sigma )

So, for our problem:

the summation of ( 3 + 2k ) as k goes from 8 to 50

n = 50 - 7 = 43
 

Back to the problem:

Sn = ( n / 2 )( a8 + a50 )
 

S43 = ( 43 / 2 )( 13 + 103 ) = 2623

 

YOUR TURN:

Find the summation of ( 4k - 7 ) as k goes from 20 to 100