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

(r + 1)th term in the given expansion is given by tr+1=10Cr 210−r2 3r5where r = 0, 1, 2, ..., 10For rational termsr = a multiple of 5 = 0, 5, 10……..(1)10 – r = a multiple of 2 = 0, 2, 4, 6, 8, 10 ……...(2)From Eq. (1) and (2) possible values of r are : 0 and 10∴ sum of rational terms =t1+t11=10C0(2)10(31/5)0+10C10(2)0(31/5)10 =25+32=32+9=41