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

1.	The no. of irrational terms in the expansion of (51/8 + 21/6)100
Answer» The number of irrational terms in the expansion is 97. Explanation: From binomial theory we know that the (r+1)th term of the expansion (a+b)ⁿ is (ⁿC_r)(a^n-r)(b^r) In this case a = 5^1/8 ; b = 2^1/6 ; 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

Answer»

The number of irrational terms in the expansion is 97.

Explanation:

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

(a+b)ⁿ is (ⁿC_r)*(a^n-r)*(b^r)

In this case a = 5^1/8 ; b = 2^1/6 ; 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

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

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