in the expansion of {1+x]43 , coefficients to {2r+1}th term and {r+2}th terms are equal . find r
 

R=1

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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​ b

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
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