D/dt integral from 0,2x of (e^t+2t)dt is?
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<h3>Answer: -n/(n+1)</h3>
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Explanation:
Let's expand this out up to say n = 6 terms
I'll write the summation in such a way that each (1/(k+1) - 1/k) grouping will get its own row. Each row has the same value of k.
If we plug in k = 1 through k = 6, but not evaluate just yet, we will have this:
1/(1+1) - 1/1
1/(2+1) - 1/2
1/(3+1) - 1/3
1/(4+1) - 1/4
1/(5+1) - 1/5
1/(6+1) - 1/6
All I've done so far is replace k with 1 through 6. Again, each row represents a different k value. Each row has the general format 1/(k+1)-1/k.
Let's simplify everything in the first column.
So the (1+1) turns into 2, the (2+1) turns into 3, and so on.
Doing that leads to...
1/2 - 1/1
1/3 - 1/2
1/4 - 1/3
1/5 - 1/4
1/6 - 1/5
1/7 - 1/6
Then note how we have the cancellations shown in the diagram below. The color coding shows how the terms pair up to cancel out. By "cancel out", I mean specifically that the fractions add up to 0. Eg: 1/2 + (-1/2) = 0.
Nearly everything cancels out. The only things left after the dust settles is the -1/1 in the first row and the 1/7 in the bottom row. This evaluates to 1/7-1/1 = 1/7 - 7/7 = (1-7)/1 = -6/7
If you were to try this with n = 7, then you should find that again nearly everything cancels but the fractions -1/1 and 1/8. Then 1/8 - 1/1 = -7/8.
For n = 8, you should end up with 1/9 - 1/1 = -8/9.
You can probably see the pattern.
A conjecture is that the answer is -n/(n+1). The general proof of this isn't too tricky. Simply follow the same ideas as mentioned above. Expand out a few terms and see how things cancel out. You'll find that everything cancels except for the -1/1 in the first row and the 1/(n+1) just one term before the very end. Then we can say:
1/(n+1) - 1/1
1/(n+1) - (n+1)/(n+1)
(1-(n+1))/(n+1)
(1-n-1)/(n+1)
-n/(n+1)
If we tried n = 6, then we find,
-n/(n+1) = -6/(6+1) = -6/7
which was the result we found earlier when we added the first n = 6 terms of this series. Trying out n = 7 should lead to -n/(n+1) = -7/8, and so on.