The number of integral terms in the expansion of (71/2+51/6)78 is
a) 15
b) 14
c) 16
d) 0
I know the ans is b. But I want the method of solving it. What is the condition for finding the no. of integral terms?
Dear Sohini,
This is a good question; a lay man approach has been presented for this problem.
You have,
(71/2 + 51/6 )78
Now, general term of this binomial will be
Tr+1 = 78Cr (71/2)78-r (51/6)r
This general term will be an integer if 78Cr is an integer and 778-r/2 is an integer and 5r/6 is an integer
78Cr will always be a +ve integer
Since 78Cr denotes no. of ways of selecting r things out of 78 things, it cannot be a fraction.
778-r/2 will be an integer
If 78-r/2 is an integer
r = 0 , 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16, ………………..70 , 72 , 74 , 76 and 78 (i)
Also 5 is an integer if r/6 is an integer
r = 0 , 6, 12 , 18 , 24 , ………………… (ii)
Taking the intersection (i) and (ii)
We have,
r = 0 , 6 , 12 , 18 , 24 , 30 , 36 , 42 , 48 , 54, 60 , 66 , 72 , 78
Hence total r permissible is 14
Therefore you have 14 terms
From binomial theory we know that the (r+1)th term of the expansion
(a+b)^n is (nCr)*(a^n-r)*(b^r)
In this case a = 5^(1/6) ; b = 2^(1/8) ; 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 6 and 8 respectively so as to cancel out the fractional exponents.
=>n-r = 6k and r = 8m where k and m are some non negative integers
=> r should take values 0,8,16,24,32,40,48,56,64,72,80,88,96 (8m)
=> n-r should take values among 0,6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 (6k)
As n= 100 for n-r=6k, r should take 100,94,88,82,76,70,64,58,52,46,40,34,28,22,16,10,4
The common values of r which satisfy r = 8m and n-r = 6k simultaneously are r = 16,40,64,88
=> There will be 4 rational terms
=> The remaining 97 terms will be irrational
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