If the coefficients of (2r+1)th  term and (r+2)th  term in the expansion of (1+x)49  are equal, then r=

Given  expansion  (1+x)49 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= 49C2r (1)49−2r x2r    T2r+1=49C2r x2r And (r+2)th  termand Tr+2= 49Cr+1 (1)49−rxr+1 Tr+2= 49Cr+1 xr+1 According to problem  the coefficients of (2r+1)th  and (r+2)th  terms are equalGiven  49C2r= 49Cr+1 (∴nC2r=nCr−1⇒2r=r−1 ,n=2r+r−1) ⇒2r=r+1  or 2r+r+1=49 ⇒                 r=1  or  r=16