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

*print*Print*list*Cite

### 2 Answers

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

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

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

`=a^4+4a^3b+4a^3c+6a^2b^2+12a^2bc+6a^2c^2+4ab^3`

`+12ab^2c+12abc^2+4ac^3+b^4+4b^3c+6b^2c^2+4bc^3+c^4`

**Sources:**

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.**