Expert: Scott A Wilson - 1/14/2010

Question
A lot of thanks and gratitude from the bottom of my heart for reading this e-mail.
Sir, a research work has been going on around the world to resolve a debate regarding a financial matter for the last more than 3 decades. I have been working extensively for the last 5 years on that matter and about to solve the problem. At the final stage of my work, I need to have summations of 2(two) geometric progressions mentioned below:

1)   summation of n number of terms of the following series:

a(1) + aq{1+ (1+2)} +  aq2{1+(1+2) +(1+2+3)}+ aq3{1+(1+2) +(1+2+3)+(1+2+3+4)} +
aqn-1{1+(1+2) +(1+2+3)+(1+2+3+4) +…+(1+2+3+…..+n)}

2) Summation of n number of terms of the following series:

a(1) + aq{1+ (1+2) +1} +  aq2[{1+(1+2) +(1+2+3)} + {1+ (1+2) +1}]+
aq3[{1+(1+2) +(1+2+3)}+(1+2+3+4)} + {1+(1+2) +(1+2+3) + 1+ (1+2) +1)}]
aqn-1[{1+(1+2) +(1+2+3)}+(1+2+3+4)} +…+(1+2+3+…..+n)}+ {1+(1+2) +(1+2+3)} + …..+   1+ (1+2) +1}]

Equation No. 2 can also be written as follows:

a(1) + aq{1+ (1+2) + (numeric number of previous term which equals to 1)} +  
aq2{1+(1+2) +(1+2+3) + (summation of numeric numbers of previous term which equals to 5)}+
aq3{1+(1+2) +(1+2+3)+(1+2+3+4) + (summation of numeric numbers of previous term which equals to 15)}+
aq4{1+(1+2) +(1+2+3)+(1+2+3+4+) +(1+2+3+4+5) + (summation of numeric numbers of previous term which equals to 35)}+
aq5{1+(1+2) +(1+2+3)+(1+2+3+4+) +(1+2+3+4+5) +(1+2+3+4+5+6) + (summation of numeric numbers of previous term which equals to 70)}+…

OR

a(1) + aq(5) +  aq2(15) + aq3(35) + aq4(70) + aq5(126) + aq6(210) + aq7(330)  +…
 
Your assistance will surely help us to resolve a issue which remains unresolved for the last 3 decades globally. Sir, Your contribution will be properly recognized.

Best regards,

Shafayet Arefin
Deputy Director
Bangladesh Bank
Phone: 01552 476764
8056473(Res)
9559672(Office)


Answer
The summation of 1+2+3+4+5 + ... N = N(N+1)/2.

The sum of the sum is 1, 3, 6, 10, 15, ...and is found by (Nģ+3Nē+2N)/6.

The sum of the sum of the sum is 1, 5, 15, 35, 70, 126, ...
and is found by using a a difference table.

I'll have to try and work that last one out tonight,
so right back again if you need it done.

Also, I need to shut down fairly soon and
haven't finished making sure I answered all the question.