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Sum of arithmetic series...Consider a finite arithmetic series which has the form: x + (x+d) + (x+2d) + (x+3d) + ... + (x+(n-1)d) where x is the initial value and d is the common difference. Find the sum of this series. We're given the hint: suppose you wanted to add up all the numbers from 1-1000. The standard trick is to write the series first in ascending order, then in descending order and then add the corresponding terms like so: First ascending: 1, 2, 3, ... 999, 1000 Then descending: 1000, 999, 998 ... 2, 1 Now add: 1+1000=1001, 2+999=1001, 3+998=1001, ... 999+2=1001, 1000+1=1001 Now notice that the sum of all the terms in the last line is easy to calculate and, it's exactly twice the quantity we seek. So: 1+2+3+ ... +999+1000= [1000(1001)/2] = 500500 Use this same method to answer the more general question. Any help would be greatly appreciated! Thank you! |

Re: Sum of arithmetic series...Quote:
nx + [0 + d + 2d + ... + (n-1)d] You can then factor out d: nx + d[0 + 1 + 2 + ... + (n-1)] Since you know how to sum 1 + 2 + ... + n, I imagine you can solve this now? |

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