The sum of rational terms in the expansion of (2+31/5)10  is

General term in the given  expansion of (2+31/5)10  is (∴  Tr+1= nCr xn−r (a)r  be the expansion of (x+a)n) Tr+1= 10Cr (2)10−r (31/5)r Tr+1= 10Cr(2)10−r2 (3)r5,  r=0,1,2,...,  10 This will be a rational number if booth 10−r2  and r5  are whole numbers 10−r2=0⇒r=10 r5=0⇒r=0 i.e. if r=0  and r=10  So, there are only two rational terms, namely T0+1  and T10+1 Sum of these terms =T1+T11                                     =  10C0 25+ 10C10 32 (∴  Tr+1= nCr xn−r (a)r                                      =32+9=41