Hard Summation Problem I'm struggling with

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Tanish Shukla

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Hi experts, as you saw from the title I have encountered a summation problem that is quite challenging for me.
Please check the question and my working attached below.
As you can see I have solved questions (a), (b), and (c) but I don't know where to start with parts (d) and (e).
Obviously the results from (a), (b) and (c) will be used in solving these, but I can't quite put the pieces together.
Could one of the experts please help me?
(Sorry if my handwriting is a bit messy)
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I'm not sure if there's much to show. It's a telescoping series. Things cancel and you are left with [imath]n^3[/imath].
[math]\sum_{r=1}^{n} \left( r^3 - (r-1)^3 \right) = (1^3-0^3)+(2^3-1^3)+(3^3-2^3)+...+(n^3-(n-1)^3)[/math]
 
BigBeachBanana said:
I'm not sure if there's much to show. It's a telescoping series. Things cancel and you are left with [imath]n^3[/imath].
[math]\sum_{r=1}^{n} \left( r^3 - (r-1)^3 \right) = (1^3-0^3)+(2^3-1^3)+(3^3-2^3)+...+(n^3-(n-1)^3)[/math]
Click to expand...
Thank you for your response.
Could you give me any hints as to how to do part (e)?
 
If I have the difference of cubes and I run into trouble, then I consider factoring. Try that and see where that gets you.
 
Steven G said:
If I have the difference of cubes and I run into trouble, then I consider factoring. Try that and see where that gets you.
Click to expand...
I tried this, but the difference of cubes method isn't really taught in this book and wouldn't really be an acceptable method in the mark scheme of an actual IGCSE test of this specification.
The difference of cubes method got me to the same place I was before: at [imath]3r^2-3r+1[/imath], and I don't really see what I can do with this. I already solved part (c), the summation of [imath]3r^2-3r+1[/imath].
 
So what is [imath]\sum_r (3r^2-3r+1)[/imath] ? How about [imath]\sum_r 3r[/imath] ?
 
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