SOLUTION: find the sum of following series: n*1+(n-1)*2+(n-2)*3+.........+n*1

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Question 174497: find the sum of following series:
n*1+(n-1)*2+(n-2)*3+.........+n*1
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

 =

sum%28+matrix%281%2C4%2C+%22%22%2C+%22%5B%22%2C+n-%28k-1%29%2C%22%5D%2Ak%22+%29%2Ck=1%2Cn%29 =

sum%28+matrix%281%2C2%2C+%22%22%2C+%28n-k%2B1%29k+%29%2Ck=1%2Cn%29 =

sum%28+matrix%281%2C2%2C+%22%22%2C+nk-k%5E2%2Bk+%29%2Ck=1%2Cn%29 =







Combine the 1st and 3rd terms

 

The first sum is the arithmetic series of the first n positive
integers,

which is given by the formula: %28n%28n%2B1%29%29%2F2

so we have

 

The remaining sum is the sum of the squares of the first n
positive integers.  This is also a well-known formula:

%28n%28n%2B1%29%282n%2B1%29%29%2F6

So now we have:

%28n%2B1%29%28%28n%28n%2B1%29%29%2F2%29+-+%28n%28n%2B1%29%282n%2B1%29%29%2F6

Rewrite the second term:

%28n%2B1%29%28%28n%28n%2B1%29%29%2F2%29+-+%28n%28n%2B1%29%2F2%29%28%282n%2B1%29%2F3%29

Factor out %28n%28n%2B1%29%29%2F2

%28%28n%28n%2B1%29%29%2F2%29%28%28n%2B1%29-%282n%2B1%29%2F3%29 

Get LCD in the right parentheses:

%28%28n%28n%2B1%29%29%2F2%29%283%28n%2B1%29%2F3-%282n%2B1%29%2F3%29

%28%28n%28n%2B1%29%29%2F2%29%28%283n%2B3%29%2F3-%282n%2B1%29%2F3%29

%28%28n%28n%2B1%29%29%2F2%29%28%283n%2B3-%282n%2B1%29%29%2F3%29%29

%28%28n%28n%2B1%29%29%2F2%29%28%283n%2B3-2n-1%29%2F3%29%29

%28%28n%28n%2B1%29%29%2F2%29%28%28n%2B2%29%2F3%29%29

%28n%2A%28n%2B1%29%28n%2B2%29%29%2F6

Edwin