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FP1 Series - Ex5F Q15 Show that when n is even...

I've been stuck on this question for a while now:

Show that when n is even:

I understand you have to split it into sum of odds - sum of evens but I don't get why you have to take away twice the sum of the even cubes from the sum of

\displaystyle \sum_{r=1}^{n}r^{3}

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Original post by saffybonbon

I've been stuck on this question for a while now:

Show that when n is even:

I understand you have to split it into sum of odds - sum of evens but I don't get why you have to take away twice the sum of the even cubes from the sum of

\displaystyle \sum_{r=1}^{n}r^{3}

Start with the sum of all the cubes:

1^3 + 2^3 + 3^3 +.... + (n-2)^3 + (n-1)^3 + n^3

Hence if we subtract the sum of even cubes once we are left with

1^3 + 3^3 +.... + (n-1)^3

And subtracting the sum of even cubes again we have

1^3 - 2^3 + 3^3 +... - (n-2)^3 + (n-1)^3 - n^3

Which is the series you wanted.

OP

Ohhhh that makes so much sense, I feel so stupid now haha. Thank you so much!!

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