# Thread: Factoring the n out of adding the areas of a definite integral problem

1. ## Factoring the n out of adding the areas of a definite integral problem

Can someone please explain how the n is factored out of the 6n and ends up with the 7n. I must learn how to factor this and not just accept the answer.
I am having to make sure my question is understood. I do not think my message will line up properly. There is an example in my text that adds the sums of the area as part of the solution of calculating the definite integral of a function.. When I get to the part of adding the sums of the aresa:
= 3/n [(7+7+….7) + (6/n + 12/n +…6n/n) ]
= 3/n [ 7n + 6/n * (1 + 2+ ….n)]
= 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

2. i don't really get what the problem is but does this help?

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

so just the last bit of the equation

6/n*n(n+1)/2

=6n(n+1)/2n

=3(n+1)

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

3. ## 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