The number of integral terms in the expansion of (7 1/2 +5 1/6 ) 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
 
						

							
What if the degree is high ? 
Like here.... Integrals solutions in expansion of (5^1/2 + 7^1/8)^1024 ? 
						

							
we can use this simple formula for finding integral co efficients ,no of rational nos...
n/L.C.M(p,q) whr p,q are the nos..eg 
lets solve tht question itself
78/lcm(2,6)=78/6=13 so to find the tot terms we do n+1 so 13+1=14 terms
another eg 1024/lcm(2,8)=128 ,tot terms =129
						

							
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