Please explain this limit problem

rishi kesh
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Please explain the above limit problem.i am able to understand last steps but can't get initial 4 steps.
 

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The sum of natural numbers 1 to N. Is that the part you are confused about?
Let's do an example. Say you want to add the numbers 1 to 10. You can group them: Take the first number and the last number (1 + 10) then 2nd number and next to last (2 + 9), then (3 + 8) etc. Each grouping adds to 11 {N+1}. There are 5 groupings {N/2}. So in general, we have (N + 1) * (N/2). This works with odd N values as well.
 
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scottdave said:
The sum of natural numbers 1 to N. Is that the part you are confused about?
Let's do an example. Say you want to add the numbers 1 to 10. You can group them: Take the first number and the last number (1 + 10) then 2nd number and next to last (2 + 9), then (3 + 8) etc. Each grouping adds to 11. There are 5 groupings. So in general, we have (N + 1) * (N/2). This works with odd N values as well.
This trick is something interesting to know about. I will remember it.appreciate your help :smile::smile:
 
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rishi kesh said:
This trick is something interesting to know about. I will remember it.appreciate your help :smile::smile:
But of you look at the 4th step they replaced 'n' by n-1 ..how do that work?
 
rishi kesh said:
But of you look at the 4th step they replaced 'n' by n-1 ..how do that work?
The comment appears to be misplaced. The replacement occurs one step above, going from ##\sum_{r=1}^n## to ##\sum_{r=1}^{n-1}##.
 
DrClaude said:
The comment appears to be misplaced. The replacement occurs one step above, going from ##\sum_{r=1}^n## to ##\sum_{r=1}^{n-1}##.
Oh i got it. :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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