# prove that W0+b1*j’*(j’+1)-b0*j”*(j”+1) = W0+b1+b0*j”+1+(b1-b0)*(j”+1)(j"+1) where J’= j”+1

*print*Print*list*Cite

### 1 Answer

You need to substitute `j'` for `j'' + 1` to the left and right side, such that:

`w_0 + b_1*j'(j' + 1) - b_0*(j' - 1)*j' = w_0 + b_1 + b_0*(j' - 1) + 1 + (b_1 - b_0)*j'*j'`

You need to open the brackets and reduce duplicate terms, such that:

`b_1*(j')^2 + b_1*j' - b_0*j' + b_0 = b_1 + b_0*j' - b_0 + 1 + b_1*(j')^2 - b_0*(j')^2`

`b_1*j' - b_0*j' + b_0 = b_1 + b_0*j' - b_0 + 1 - b_0*(j')^2`

`b_1*j'+ 2b_0 = b_1 + 2b_0*j' + 1 - b_0*(j')^2`

`b_1(j' - 1) -2b_0(j' - 1) = 1 - b_0*(j')^2`

`(j' - 1)(b_1 - 2b_0) = 1 - b_0*(j')^2`

**Hence, testing the identity, using the provided condition `j' = j'' + 1` , yields that the identity does not hold.**