If in the expansion of (1+x)43 , the coefficients of (2r+1)th  and (r+2)th  terms are equal, then r  is equal to

Given  expansion  (1+x)43 is, We have general term in the expansion (x+a)n (∴  Tr+1= nCr xn−r (a)r  be the expansion of (x+a)n)  (2r+1)th  termT2r+1= 43C2r (1)43−2r x2r    T2r+1=43C2r x2r And (r+2)th  termand Tr+2= 43Cr+1 (1)43−rxr+1 Tr+2= 43Cr+1 xr+1 According to problem  the coefficients of (2r+1)th  and (r+2)th  terms are equalGiven  43C2r= 43Cr+1 (∴nC2r=nCr−1⇒2r=r−1 ,n=2r+r−1) ⇒2r=r+1  or 2r+r+1=43 ⇒  r=1  or 14 . (∵C(n,r))=C(n,n−r)