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Binomial theorem for positive integral Index

Question

The number of irrational terms in the expansion of (58+26)100, is

Moderate
Solution

We have, 

          (58+26)100=r=0100100Cr (58)100r(26)r (58+26)100=r=0100100Cr 5100rr8=r=0100Tr+1

         where Tr+1=100Cr5100r82r6

   Clearly  Tr+1 will be an integer if 100r8 and r6 are integers.

 This is possible when 100r is a multiple of 8 and r is a multiple 

 of 6 

 100r=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.



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