adding sums, definite integral

I am not sure my clarification to my question came through, so I am posting it again.

http://www.mathhelpforum.com/math-he...cons/icon1.gif **White Noise, let me clarify**

= 3/n [ 7 + 6/n = 7 + 12/n + ....+ 7 + 6n/n] goes to

= 3/n [ ( 7 + 7 + 7) + (6/n + 12/n + ....+ 6n/n)] which goes to

= 3/n[ 7n + 6/n (1 + 2 +...+n)] which goes to

= 3/n[ 7n + 6/n * n(n+1)/2 ]

= 3/n [7n + 3((n+1)]

= 3/n[ 7n + 3n + 3]

= 3/n [10n + 3]

= 30 + 9/n I do not understand the algebra. I do not see how the term 6n/n was factored out and became 7n + 6/n(1+2+....+n) Can you please explain the algebra that accomplishes this? Thanks

Mr. Jameson. I apologize for the double posting in regards to my question regarding a

You are correct. The = sign is a mistake, it should have been a + sign. I was too nervous being unable to understand how the n was factored out and ended up with the 7, making the term 7n. I thought I was alright in regards to factoring but I was unable to understand how to add the sums of the areas with part of the steps, to get the proper equation to get the limit of the function, the definite integral.

After I studied a few more examples in my text, I now think that factoring the n out of the last right hand value and multiplying it by the 7 is simply a rule. Something to be done every time sums of areas are used. In this case factoring out the n to arrive at

= 3/n [(7+7+...7) + (6/n + 12/n + ...6n/n)]

= 3/n [7n + 6/n (1+2+...n)]

the 7n must be the result of a "rule" in regards to adding sums of areas

= 3/n [7n + 6/n * n(n+1)/2]

= 3/n [7n+ 3(n+1)]

= 3/n [7n + 3n + 3]

= 3/n [ 10n + 3]

= 30 + 9/n

Take the limit as n --> infinity

5

ò (2x + 3 ) dx = lim ( 30 + 9/n) = 30

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