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I'm having some trouble finding the best approach to the following arithmetic sequence problem:

$$\sum_{i=3}^{30}{[(i-3)^2+i-3]}$$

I'm aware that I can break up the sequence:

$$\sum_{i=3}^{30}{[(i-3)^2+i-3]} = \sum_{i=3}^{30}{(i-3)^2} + \sum_{i=3}^{30}{i} - \sum_{i=3}^{30}{3}$$

However, I'm stuck on the identity for the first term: $\sum_{i=3}^{30}{(i-3)^2}$

Given that the summation index is $i=3$, I cannot find an identity for a quadratic arithmetic sequence where $i \ne 1$

Any advice would be appreciated.

thisisanon
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2 Answers2

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Set $i-3=j,$ $$\sum_{i=3}^{30}(i-3)^2=\sum_{j=0}^{27}j^2=\sum_{j=1}^{27}j^2$$

Now see How to get to the formula for the sum of squares of first n numbers?

Community
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lab bhattacharjee
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  • Is it really as simple as that? I tried working out some arithmetic sequences where $i \ne 1$ and the answer didn't seem to correlate if I set the summation index to 1. But I'll try it - thanks! – thisisanon Oct 05 '14 at 16:20
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    @ptikobj, For $$\sum_{r+1}^nk^2$$ use $$\sum_{r+1}^nk^2=\sum_1^nk^2-\sum_1^rk^2$$ – lab bhattacharjee Oct 05 '14 at 16:22
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First of all, we have$$\sum_{i=3}^{30}((i-3)^2+(i-3))=\sum_{k=1}^{27}(k^2+k)=\sum_{k=1}^{27}k(k+1).$$ Here, noting that $$k(k+1)=\frac 13\left((k+2)(k+1)k-(k+1)k(k-1)\right),$$ we have $$\sum_{k=1}^{27}k(k+1)=\frac 13\sum_{k=1}^{27}((k+2)(k+1)k-(k+1)k(k-1))$$$$=\frac 13((3\cdot 2\cdot 1-0)+(4\cdot 3\cdot 2-3\cdot 2\cdot 1)+\cdots$$$$+(28\cdot 27\cdot 26-27\cdot 26\cdot 25)+(29\cdot 28\cdot 27-28\cdot 27\cdot 26))$$$$=\frac{29\cdot 28\cdot 27}{3}=7308.$$

mathlove
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