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Find Formula (Posted on 2012-02-17) Difficulty: 3 of 5
Derive a formula for the product of the first n members of an arithmetic progression with an initial element a and a common difference d;(a*d>0).
Verify the validity of your expression for (a,d,n)=(2,3,8).

See The Solution Submitted by Ady TZIDON    
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possible solution | Comment 1 of 4

As we will be taking sums of numerous products, some definitions will be in order:

Let A be the index of the series. Let n=(A-1), for convenience, since the variable d will not appear in the first term of any member of the series.

Let k be a number of integers to be multiplied together (k=2 for two integers, k=3 for 3 integers etc.) Their product is P.                                         

The expression PC(n,2) then means choose each distinct pair of numbers from 1 to n and take their products, and PC(n,k)  means choose each distinct combination of k numbers and take their products.                                                

The total of all such multiplications is then summed: ÓPC(n,k)   

Now we have: a^(n+1)+(ÓPC(n,1))a^(n)d+(ÓPC(n,2))a^(n-1)d^2+ (ÓPC(n,3))a^(n-2)d^3...


Since in the last term the numbers from 1 to d are multiplied together once, that term is a factorial.

As redefined the problem now reads, 'Verify the validity of your expression for (a,d,A)=(2,3,8).'


2^(7+1)+(ÓPC(7,1))2^7*3+(ÓPC(7,2))2^(7-1)3^2+(ÓPC(7,3))2^(7-2)3^3+(ÓPC(7,4))2^(7-3)3^4 +(ÓPC(7,5))2^(7-4)3^5+(ÓPC(7,6))2^(7-5)3^6+2*7!*3^7.

Evaluating this gives a^8+28a^7d+322a^6d^2+1960a^5d^3, whose coefficients correspond to A094638 in Sloane, so then, using the table: 6769a^4d^4+13132a^3d^5+13068a^2d^6+5040ad^7.

Substituting: 2^8+28*2^7*3+322*2^6*3^2+1960*2^5*3^3+6769*2^4*3^4 +13132*2^3*3^5+13068*2^2*3^6+2*5040*3^7 = 96342400.

By direct calculation:
2*5*8*11*14*17*20*23 = 96342400


Edited on February 18, 2012, 12:43 am
  Posted by broll on 2012-02-18 00:37:30

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