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

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 C_r * a^n-r​ b^r 

here term=(2r+1)
so r in the above formula will be: 2r 
Now expand using 2r and we get     ⁿC_2r *1^43-2r​ *x^2r
now doing the same for the r+2^th term we have ⁿC_r+1 * 1 ^43-r+1 * x^r+1

Note  1ⁿ =1 (meaning 1 raise to any number is 1)or(1*1*1*1*1......n=1) 
so we know 1^43-2r​ and 1^43-r+1 are both equal to one.
Given the 2r+1 and r+2 terms are equal

we can equate them

ⁿC_2r *1^43-2r​ *x^2r   =​ ⁿC_r+1 * 1 ^43-r+1 * x^{r+1    
n}C_2r​ *x^2r   =​ ⁿC_r+1* x^r+1
Now we know the powers of x will be as the terms are equal

hence we equate the x terms(without coefficients)
x^2r  =​x^{r+1   as they have the same base when  x divides  the exponents will only subtract (}x^2r /​x^r+1)

so x^2r-r=x¹
hence x^r=x¹
and r=1