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Summation starting from 0

Please could someone explain how to deal with this summation from 0? I don't understand the mark scheme...

Thanks!
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Reply 1
Avatar for Zacken
Zacken
22
Original post by PhyM23
Please could someone explain how to deal with this summation from 0? I don't understand the mark scheme...

Thanks!


Sure. ∑r=0nf(r)=f(0)+f(1)+⋯+f(n)=f(0)+∑r=1nf(r)\displaystyle \sum_{r=0}^n f(r) = f(0) + f(1) + \cdots + f(n) = f(0) + \sum_{r=1}^n f(r).

In your case:

Unparseable latex formula:

\displaystyle[br]\begin{align*} \sum_{r=0}^n(r^2 - 2r + 2n +1) &= (0^2 - 2(0) + 2n + 1) + \sum_{r=1}^n (r^2 - 2r + 2n + 1) \\ &= (2n+1) + \sum_{r=1}^n (r^2) - 2\sum_{r=1}^n (r) + (2n+1)(n) \\ &= (2n+1)(n+1) + \sum_{r=1}^n r^2 - 2\sum_{r=1}^n r \\ & = \cdots\end{equation*}

(edited 8 years ago)
Original post by PhyM23
Please could someone explain how to deal with this summation from 0? I don't understand the mark scheme...

Thanks!


Well what part of the sum is affected by r=0
Reply 3
Avatar for Zacken
Zacken
22
Original post by PhyM23
Please could someone explain how to deal with this summation from 0? I don't understand the mark scheme...

Thanks!


Oh, and it's worth bearing in mind that nn is just a constant, so: ∑r=0n(2n+1)=(2n+1)(n+1)\displaystyle \sum_{r=0}^n (2n+1) = (2n+1)(n+1) - i.e: you're adding the number (2n+1)(2n+1) to itself (n+1)(n+1) times.
Reply 4
Avatar for φM23
φM23
OP
7
Original post by Zacken
Oh, and it's worth bearing in mind that nn is just a constant, so: ∑r=0n(2n+1)=(2n+1)(n+1)\displaystyle \sum_{r=0}^n (2n+1) = (2n+1)(n+1) - i.e: you're adding the number (2n+1)(2n+1) to itself (n+1)(n+1) times.


It makes perfect sense now. Thank you! :smile:

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