The number of irrational terms in the expansion of (58+26)100, is
We have,
(58+26)100=∑r=0100 100Cr (58)100−r(26)r⇒ (58+26)100=∑r=0100 100Cr 5100−rr8=∑r=0100 Tr+1
where Tr+1=100Cr5100−r82r6
Clearly Tr+1 will be an integer if 100−r8 and r6 are integers.
This is possible when 100−r is a multiple of 8 and r is a multiple
of 6
⇒ 100−r=0,8,16,….,96 and ,r=0,6,12,…,96
⇒ r=4,12,20,…,100 and r=0,6,12,….,96
⇒ r=12, 36, 60, 84
Hence, there are 4 rational terms and 101 - 4 = 97 irrational
terms.