Given (1+x)^43 is the binomial term :
and terms in the binomial expansion are r+2 and 2r+1
using the General formula Tr+1=n Cr * an-r br
here term=(2r+1)
so r in the above formula will be: 2r
Now expand using 2r and we get nC2r *143-2r *x2r
now doing the same for the r+2th term we have nCr+1 * 1 43-r+1 * xr+1
Note 1n =1 (meaning 1 raise to any number is 1)or(1*1*1*1*1......n=1)
so we know 143-2r and 143-r+1 are both equal to one.
Given the 2r+1 and r+2 terms are equal
we can equate them
nC2r *143-2r *x2r = nCr+1 * 1 43-r+1 * xr+1
nC2r *x2r = nCr+1* xr+1
Now we know the powers of x will be as the terms are equal
hence we equate the x terms(without coefficients)
x2r =xr+1 as they have the same base when x divides the exponents will only subtract (x2r /xr+1)
so x2r-r=x1
hence xr=x1
and r=1