(a1+a2+a3+...........+ax)^n = ?? where, n and x are integer numbers and a1,a2,a3,.....ax are real numbers     

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

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We need an expression for `(a_1+a_2+...a_k)^n` where `k,n in NN,a_i in RR` .

The required expression is

`(a_1+a_2+...+a_k)^n=sum_(j=0)^n a^j(a_2+a_3+...a_k)^(n-j)(n/j)` where `(n/j)` is meant to be the binomial coefficient.

Small example: `(a+b+c)^4=sum_(j=0)^4 a^j(b+c)^(n-j) (n/j)`

`=a^0(b+c)^4(4/0) + a^1(b+c)^3(4/1) + a^2(b+c)^2(4/2)+a^3(b+c)^1(4/3)+a^4(b+c)^0(4/4)`


`=b^4+ 4b^3c+6b^2c^2+4bc^3+c^4+4a(b^3+3b^2c+3bc^2+c^3)`




sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should consider several possibilities with respect to the terms of the given sum.

`a_1 = 1, a_2 = 2,.....,a_x = x=gta_1 + a_2 + ... + a_x = 1+2+...+x = x*(x+1)/2`

Hence, the n th power of this sum is: `(x^n)*(x+1)^n/2^n`

`` If the series a_1, a_2,...,a_n is arithmetical progression then `(a_1 + a_2 + ... + a_x)^n = (x*(a_1 + a_x))^n/2^n`

If the series `a_1, a_2,...,a_n`  is geometric progression then `(a_1 + a_2 + ... + a_x)^n = (a_1*(q^x - 1))^n/(q-1)^n` , q denotes the common ratio.

Hence, considering several relations between the terms of the addition, the sum may be: `s = (x^n)*(x+1)^n/2^n; s =(x*(a_1 + a_x))^n/2^n; s = (a_1*(q^x - 1))^n/(q-1)^n` , nevertheless the number of possible relations between the terms is countless.

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