MHB Double summation: inner index = function of outer index

[hitanshu_sachania]

Here N, a, and b are integer constants. M is also an integer but changes for every value of x, which makes the index of the second summation dependent on the first. The problem is the relationship M(x) is analytically difficult to define. Is there a way to solve/simplify this expression?

 

[skeeter]

https://mathforums.com/threads/doub...x-function-of-outer-index.350763/#post-629045

 

[HOI]

There must be a relation between y and x according to[ which as the value of x varies, y will vary, so would M(x).

No, the only "relation" between y and x is the stated one- that y goes from 1 to M(x). For example,
$\sum_{x= 1}^3\sum_{y= 1}^{x+ 1} F(x, y)$ where "M(x)" is "x+ 1".

For x= 1 y goes from 1 to 2- the inner sum is F(1, 1)+ F(1, 2).
For x= 2 y goes from 1 to 3- the inner sum is F(2, 1)+ F(2, 2)+ F(2, 3).
For x= 3 y goes from 1 to 4- the inner sum is F(3, 1)+ F(3, 2)+ F(3, 3)+ F(3, 4).
$\sum_{x= 1}^3\sum_{y= 1}^{x+ 1} F(x, y)$= F(1, 1)+ F(1, 2)+ F(2, 1)+ F(2, 2)+ F(2, 3)+ F(3, 1)+ F(3, 2)+ F(3, 3)+ F(3, 4).