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$$1 (N) + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \cdots + N (1) $$

I need to write it in simplest form?

here 1(N) means 1 multiply by N

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marked as duplicate by Martin Sleziak, Najib Idrissi, RecklessReckoner, 6005, jkabrg Dec 16 '15 at 15:31

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Sorry, but does $N$ stand for a complex number? – S. Snape Nov 29 '12 at 10:05
    
Babak, $N$ must be a natural number; $i$ is intended as an index, not the square root of $-1$. So the sum is $\sum_{k=0}^{N-1} (1+k)(N-k)$. (Ah, now I see why you asked that. Grijesh, the tags you've chosen are completely inappropriate.) – Rhys Nov 29 '12 at 10:12
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I am tiring to answer some other problem related to computer scince – Grijesh Chauhan Nov 29 '12 at 10:17
    
Apart from the post already mentioned by Normal Human, some other posts which are linked there might be interesting for you, too. – Martin Sleziak Dec 16 '15 at 14:12
up vote 7 down vote accepted

$$ \sum_{k=1}^Nk(N-k+1)=\sum_{k=1}^N kN-\sum_{k=1}^Nk^2+\sum_{k=1}^Nk=N\,\frac{N(N+1)}2-\frac{N(N+1)(2N+1)}6+\frac{N(N+1)}2=\frac{N(N+1)}2\left(N-\frac{2N+1}3+1\right)=\frac{N(N+1)(N+2)}6. $$

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There is a typo. It should be N+2 in place of N+4. – Gautam Shenoy Nov 29 '12 at 10:49
    
@Gautam Shenoy - thanks , Martin Argerami - thanks! – Grijesh Chauhan Nov 29 '12 at 10:52
    
@draks 10/6 = 1 floor function – Grijesh Chauhan Nov 29 '12 at 12:20
    
how are you expanding $\sum_{k=1}^N kN$ to $N\,\frac{N(N+1)}2$ ? – david_adler Jun 13 '15 at 16:29
    
It's the basic identity $\sum_{k=1}^Nk=\frac{N(N+1)}2$, multiplied by $N$. – Martin Argerami Jun 13 '15 at 17:37

Your sum can be written as

$$\sum_{k=1}^Nk(N-k+1) = N\sum_{k=1}^{N}k - \sum_{k=1}^N{k^2} + \sum_{k=1}^N{k}$$

$$ = \frac{N(N+1)^2}{2} - \frac{N(N+1)(2N+1)}{6}$$

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thank Gautam Shenoy ! – Grijesh Chauhan Nov 29 '12 at 10:53

A more complex but fun alternative solution using Finite Calculus:

$$ \begin{align}  \sum_{x=1}^N x \left[(N+1)-x\right] & = \sum_{1}^{N+1} \left[ Nx - x(x-1)\right] \delta x = \sum_{1}^{N+1} \left( Nx^{(1)} - x^{(2)} \right) \delta x \\ & = \left[ \frac{N}{2}x^{(2)} - \frac{1}{3} x^{(3)} \right]_{1}^{N+1} = \left[ \frac{N}{2}x(x-1) - \frac{1}{3} x(x-1)(x-2) \right]_{1}^{N+1}\\ & = \left[ \frac{N(N+1)^2}{2} - 0 \right] - \left[ \frac{1}{3}(N+1)N(N-1) - 0 \right] \\ & = \frac{N(N+1)(N+2)}{6}. \end{align} $$

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@ffrenz : Thanks frenz!! – Grijesh Chauhan Nov 29 '12 at 12:41

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