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The no. of irrational terms in the expansion of (51/8 + 21/6)100

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The number of irrational terms in the expansion is 97.


From binomial theory we know that the (r+1)th term of the expansion

(a+b)n is (nCr)*(an-r)*(br)

In this case a = 51/8 ; b = 21/6 ; n=100 ;

=> Total number of terms = n+1 = 101

For a term to be rational the powers of ‘a’ and ‘b’ should be integral multiples of 8 and 6 respectively so as to cancel out the fractional exponents.

n-r = 8k and r = 6m where k and m are some non negative integers
n-r should take values 0,8,16,24,32,40,48,56,64,72,80,88,96 (8k)
r should take values among 0,6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 (6m)

As n= 100 for n-r = 8k, r should take 100,92,84,76,68,60, 52,44, 36,28,20,12,4

The common values of r which satisfy r = 6m and n-r = 8k simultaneously are r = 12, 36, 60, 84.

There will be 4 rational terms thus it is obvious that remaining 97 terms will be irrational

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