document.write( "Question 603123: What is the sum of an arithmetic series with Sigma (n on top, k=50 on the bottom, and the equation is just k.) This should be easy but I'm just having a major brain fart today... \n" ); document.write( "

Algebra.Com's Answer #380621 by KMST(5328) $\"\"$ $\"About$

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$\"S=sum%28k%2Ck=50%2Cn%29\"$ =50+51+52+ ... +(n-2)+(n-1)+n is a sum of n-49 terms.
\n" ); document.write( "It can also be written as:
\n" ); document.write( " $\"S=sum%28k%2C1%2Cn%29-sum%28k%2C1%2C49%29\"$ = $\"sum%28k%2C0%2Cn%29-sum%28k%2C0%2C49%29\"$
\n" ); document.write( "There are many ways to approach the problem using \"pre-cooked\" formulas and results.
\n" ); document.write( "Otherwise, cooking from scratch,
\n" ); document.write( " $\"2%2AS=S%2BS\"$ =50+51+52+ ... +(n-2)+(n-1)+n+50+51+52+ ... +(n-2)+(n-1)+n, with 2(n-49) terms.
\n" ); document.write( "Pairing terms (associative property) \"head to tail\" (first with last, second with second last, and so on), we get the sum of n-49 pair sums (n-49 sums of two numbersin brackets)
\n" ); document.write( "2S=(50+n)+(51+(n-1))+(52+(n-2)+ ...=50+50+50+ ... = 50(n-49)
\n" ); document.write( "So, if $\"2%2AS=50%28n-49%29\"$ ---> $\"S=highlight%2825%28n-49%29%29\"$
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